University  of  California  •  Berkeley 


THE  THEODORE  P.  HILL  COLLECTION 

°f 

EARLY  AMERICAN  MATHEMATICS  BOOKS 


ELEMENTS 


GEOMETRY 


CONIC   SECTIONS. 


BY  ELIAS   LOOMIS,  LUX, 

PROFESSOR    OF   NATUEAL   rHILOBOPHT  AND    ASTRONOMY    IN   TALE    COLLEGE,    AM>    ATTIIOS    OP 
A  "COCE8E   OF   MATHEMATICS" 


TWENTY-EIGHTH  EDITION. 


NEW    YORK: 

HARPER    &    BROTHERS,   PUBLISHERS 
329    &    331   PEARL    STREET, 

(FRANKLIN    SQUARE) 


Entered,  according  to  Act  of  Congress,  in  the  year  1858,  b} 

ELIAS  LOOMIS, 
In  (he  Clerk's  Office  of  the  Southern  District  of  New  Ycrk 


TO   THE 


UON   THEODORE  FRELINGHUYSEN,  LL,D, 

CHANCELLOR   OF   THK   UNIVERSITY  OF  THE   CITY   OF   NEW  YORK. 

THE    FRIEND    OF    EDUCATION,    THE    PATRIOT    STATESMAN, 
AND    THE    CHRISTIAN    PHILANTHROPIST, 


SWorfc 


IS    RESPECTFULLY    DEDICATED 


BY' 


THE   AUTHOR 


PREFACE. 

IN  the  folio  wi.g  treatise,  an  attempt  has  been  mafle  to  combine  the 
peculiar  excellencies  of  Euclid  and  Legendre.  The  Elements  of  Euc.li 
have  long  been  celebrated  as  furnishing  the  most  finished  specimens  01 
logic  ;  and  on  this  account  they  still  retain  their  place  in  many  seminariea 
of  education,  notwithstanding  the  advances  which  science  has  made  in 
modern  times.  But  the  deficiencies  of  Euclid,  particularly  in  Solid  Gf» 
ometry,  are  now  so  palpable,  that  few  institutions  are  content  with  a 
simple  translation  from  the  original  Greek.  The  edition  of  Euclid 
chiefly  used  in  this  country,  is  that  of  Professor  Playfair,  who  has  sought, 
by  additions  and  supplements,  to  accommodate  the  Elements  of  Euclid 
to  the  present  state  of  the  mathematical  sciences.  But,  even  with  these 
additions,  the  work  is  incomplete  on  Solids,  and  is  very  deficient  on 
Spherical  Geometry.  Moreover,  the  additions  are  often  incongruous 
with  the  original  text ;  so  that  most  of  those  who  adhere  to  the  use  of 
Playfair's  Euclid,  will  admit  that  something  is  still  wanting  to  a  perfect 
treatise.  At  most  of  our  colleges,  the  work  of  Euclid  has  been  super- 
seded by  that  of  Legendre.  It  seems  superfluous  to  undertake  a  defense 
of  Legendre's  Geometry,  when  its  merits  are  so  generally  appreciated 
No  one  can  doubt  that,  in  respect  of  comprehensiveness  and  scientific 
arrangement,  it  is  a  great  improvement  upon  the  Elements  of  Euclid. 
Nevertheless,  it  should  ever  be  borne  in  mind  that,  with  most  students 
in  our  colleges,  the  ultimate  object  is  not  to  make  profound  mathemati 
cians,  but  to  make  good  reasoners  on  ordinary  subjects.  In  order  to 
secure  this  advantage,  the  learner  should  be  trained,  not  merely  to  give 
the  outline  of  a  demonstration,  but  to  state  every  part  of  the  argument 
with  minuteness  and  in  its  natural  order.  Now,  although  the  model  of 
Legendre  is,  for  the  most  part,  excellent,  his  demonstrations  are 
often  mere  skeletons.  They  contain,  indeed,  the  essential  part  of  an 
argument ;  but  the  general  student  does  not  derive  from  them  the  high 
est  benefit  which  may  accrue  from  the  study  of  Geometry  as  an  exercise 
in  reasoning. 

While,  then,  in  the  following  treatise,  I  have,  for  the  most  part,  fol 
owed  the  arrangement  of  Legendre,  I  have  aimed  to  give  his  demonstra 
tions  somewhat  more  of  the  logical  method  of  Euclid.  I  have  also  made 


VJ  PREFACE, 

gome  changes  in  arrangement,     feeveral  of  Legend  re's  propositions  have 
been  degraded  to  the  rank  of  corollaries,  while  some  of  his  corollaries. 

ri  scholiums  have  been  elevated  to  the  dignity  of  primary  propositions 
»lis  lemmas  have  been  proscribed  entirely,  and  most  of  his  scholiums 
have  received  the  more  appropriate  title  of  corollary.  The  quadrature 
.4  the  circle  is  developed  in  an  order  somewhat  different  from  any  thing 
I  have  elsewhere  seen.  The  propositions  are  all  enunciated  in  general 
terms,  with  the  utmost  brevity  which  is  consistent  with  clearness  ;  and, 
in  order  to  remind  the  student  to  conclude  his  recitation  with  the  enun 
ciation  of  the  proposition,  the  leading  words  are  repeated  at  the  close  ot 
each  demonstration.  As  the  time  given  to  mathematics  in  our  colleges 
is  limited,  and  a  variety  of  subjects  demand  attention,  no  attempt  has 
been  made  to  render  this  a  complete  record  of  all  the  known  propositions 
of  Geometry.  On  the  contrary,  nearly  every  thing  has  been  excluded 
which  is  not  essential  to  the  student's  progress  through  the  subsequent 
parts  of  his  mathematical  course. 

Considerable  attention  has  been  given  to  the  construction  of  the  dia 
grams.  I  have  aimed  to  reduce  them  all  to  nearly  uniform  dimensions, 
and  to  make  them  tolerable  approximations  to  the  objects  they  were  de 
signed  to  represent.  I  have  made  free  use  of  dotted  lines.  Generally, 
the  black  lines  are  nsed  to  represent  those  parts  of  a  figure  which  aro 
directly  involved  in  the  statement  of  the  proposition  ;  while  the  dotted 
lines  exhibit  the  parts  which  are  added  for  the  purposes  of  demonstration. 
In  Solid  Geometry  the  dotted  lines  commonly  denote  the  parts  which 
would  be  concealed  by  an  opaque  solid  ;  while  in  a  few  cases,  for  pecul- 
iar reasons,  both  of  these  rules  have  been  departed  from.  Throughout 
Solid  Geometry  the  figures  have  generally  been  shaded,  which  addition, 
it  is  hoped,  will  obviate  some  of  the  difficulties  of  which  students  frequent- 
ly complain. 

The  short  treatise  on  the  Conic  Sections  appended  to  this  volume  is 
designed  particularly  for  those  who  have  not  time  or  inclination  for  the 
fltudy  of  Analytical  Geometry.  Some  acquaintance  with  the  properties 
of  the  Ellipse  and  Parabola  is  indispensable  as  a  preparation  for  the 
study  of  Mechanics  and  Astronomy.  Those  who  pursue  the  study  of 
Analytical  Geometry  can  omit  this  treatise  on  the  Conic  Sections  if  it 
should  be  thought  desirable.  It  is  believed,  however,  that  some  knowl- 
edge of  the  properties  of  these  curves,  derived  from  geometrical  meth- 
ods, forms  an  excellent  preparation  for  the  Algebraical  and  more  general 
processes  of  Analytical  Geometry- 


CONTENTS 


PLANE   GEOMETRY. 

BOOK  I.  fift 

General  Principles   .        .        .        ... 

BOOK   II. 
Ratio  and  Proportion • 

BOOK  III. 
The  Circle,  and  the  Measure  of  Angles •    •** 

BOOK  IV. 
The  Proportions  of  Figures      .  i7 

BOOK  V. 
Problems  relating  to  the  preceding  Books        ....        ,        .    82 

BOOK  VI. 
Regular  Polygons,  and  the  Area  of  the  Circle W 


SOLID  GEOMETRT 
BOOK   VIL 

Planes  and  Solid  Angles 11* 

BOOK  VIIL 

PoJyedrons 127 

BOOK  IX 

Spherical  Geometry 148 

BOOK   X. 
Tbe  Three  round  Bodies  .  .  .        .  HW 


CONIC  SECTIONS. 

Parabola  .        . )77 

Ellipse 188 

Hyperbola 205 


A*./?. — When  reference  is  made  to  a  Proposition  in  the  same  Book,  only  the 
number  of  the  Proposition  is  given ;  but  when  the  Proportion  is  found  in  a 
different  Book,  the  number  of  the  Book  is  also  specified 


ELEMENTS  OF  GEOMETRY, 


BOOK  I. 

GENERAL   PRINCIPLES. 
Definitions. 

1.  GEOMETRY  is  that  branch  of  Mathematics  which  treats 
of  the  properties  of  extension  and  figure. 

Extension  has  three  dimensions,  length,  breadth,  and  thick 
ness. 

2.  A   line  is  that  which  has  length,  without  breadth  01 
thickness. 

The  extremities  of  a  line  are  called  points.    A  point,  there- 
fore, has  position,  but  not  magnitude. 

3.  A  straight  line  is  the  shortest  path  from  one  point  to 
another. 

4.  Every  line  which  is  neither  a  straight  line,  nor  compo 
sed  of  straight  lines,  is  a  curved  line. 

E 

Thus,  AB  is  a  straight  line,  ACDB  is  a 
broken  line,  or  one  composed  of  straight 
lines,  and  AEB  is  a  curved  line. 

~D 

5.  A  surface  is  that  which  has  length  and  breadth,  without 
thickness. 

6.  A  plane  is  a  surface  in  which  any  two  points  being  ta- 
ken, the  straight  line  which  joins  them  lies  wholly  in  that  sur- 
face . 

7.  Every  surface  which  is  neither  a  plane,  nor  composed 
of  plane  surfaces,  is  a  curved  surface. 

8.  A  solid  is  that  which  has'  leiigth,  breadth,  and  thick- 
ness, and  therefore  combines  the  three  dimensions  of  exten- 
sion. 

9.  When  two  straight  lines  meet  together,  their  inclina- 


10  GEOMETRY. 

tion,  or  opening,  is  called  an  angle.  The  point  of  meeting 
:.s  called  the  vertex,  and  the  lines  are  called  Ihe  sides  of  the 
angle. 

If  there  is  only  one  angle  at  a  point,  it  may 
oe  denoted  by  a  letter  placed  at  the  vertex,  as 
the  angle  at  A. 

But  if  several  angles  are  at  one  point,  any  one  of  them  is 
expressed  by  three  letters,  of  which  the  middle  one  is  the  let 
ter  at  the  vertex. 

B 

Thus,  the  angle  which  is  contained  by  the 
straight  lines  BC,  CD,  is  called  the  angle 
BCD,  or  DCB. 

c 

Angles,  like  other  quantities,  may  be  added,  subtracted,, 
multiplied,  or  divided.  Thus,  the  angle  BCD  is  the  sum  of 
the  two  angles  BCE,  ECD ;  and  the  angle  ECD  is  the  differ 
ence  between  the  two  angles  BCD,  BCE. 

10.  When  a  straight  line,  meeting  another  straight  line 
makes  the  adjacent  angles  equal  to  one  another, 
each  of  .them  is  called  a  right  angle,  and  the 
straight  line  which  meets  the  other  is  called  a 
perpendicular  to  it. 


11.  An  acute  angle  is  one  which  is  less  than  a 
right  angle. 


An  obtuse  angle  is  one  which  :s  greater 
than  a  right  angle. 


12.  Parallel  straight  lines  are  such  as  are  — 

in  tha  same  plane,  and  which,  being  produced 

ever  so  far  both  ways,  do  not  meet. 

13.  A  plane  figure  is  a  plane  terminated  on  all  sides  by 
.ines  either  straight  or  curved. 

If  the  lines  are  straight,  the  space  they  in- 
close is  called  a  rectilineal  figure,  or  polygon, 
and  the  lines  themselves,  taken  together,  form 
the  perimeter  of  the  polygon. 


14.  The  polygon  of  three  sides  is  the  simples  of  all,  and  is 
called  a  triangle ;  that  -of  four  sides  is  called  a  quadrilateral : 
that  of  five,  a  pentagon  ;  that  of  six,  a  hexagon,  &c. 


BOOK 


n 


15.  An  equilateral  triangle  is  one  which  has  its 
three  sides  equal.  /  v 


An  isosceles  triangle  is  that  which  has  only  two 
sides  equal. 


A  scalene  triangle  is  one  which  has  three  un- 
equal sides. 

16.  A  right-angled  triangle  is  one  which  has 
a  right  angle.     The  side  opposite  the  right  an- 
gle is  called  the  hypothenuse. 

An  obtuse-angled  triangle  is  one  which  has  an  obtuse  an 
gle.  An  acute-angled  triangle  is  one  which  has  three  acute 
angles. 

17.  Of  quadrilaterals,  a  square  is  that  which  has 
all  its  sides  equal,  and  its  angles  right  angles. 


A  rectangle  is  that  which  has  all  its  angles  right  [ 
angles,  but  all  its  sides  are  not  necessarily  equal. 


A  rhombus  is  that  which  has  all   its   sides 
equal,  but  its  angles  are  not  right  angles. 

A  parallelogram  is  that  which  has  its  op- 
posite sides  parallel.  *_ 


A  trapezoid  is  that  which  has  only  two  sides    /  \ 

rallel.  /  \ 


parallel 


18.  The  diagonal  of  a  figure  is   a  line    B 
which  joins  the  vertices  of  two  angles  not 
adjacent  to  each  other. 

Thus,  AC,  AD,  AE  are  diagonals. 

19.  An  equilateral  polygon  is  one  which  has  all  its  sides 
equal.     An  equiangular  polygon  is  one  which  has  all  its  an- 
gles equal. 

20.  Two   polygons   are   mutually   equilateral  when   they 
have  all  the  sides  of  the  one  equal  to  the  corresponding  sides 
of  the  other,  each  to  each,  and  arranged  in  the  same  order. 

Two  polygons  are  mutually  equiangular  when  they  have 


IX  GEOMETKY. 

all  the  ang  es  of  the  one  equal  to  the  corresponding  anglei 
of  the  other,  each  to  each,  and  arranged  in  the  same  order. 

In  both  cases,  the  equal  sides,  or  the  equal  angles,  are  call- 
ed  homologous  sides  or  angles. 

21.  An  axiom  is  a  self-evident  truth. 

22.  A  theorem  is  a  truth  which  becomes  evident  oy  a  train 
of  reasoning  called  a  demonstration. 

A  direct  demonstration  proceeds  from  the  premises  by  a 
regular  deduction. 

An  indirect  demonstration  shows  that  any  supposition  con- 
trary to  the  truth  advanced,  necessarily  leads  to  an  absurd- 
ity. 

23.  A  problem  is  a  question  proposed  which  requires  a  so 
lution. 

24.  A  postulate  requires  us  to  admit  the  possibility  of  an 
operation. 

25.  A  proposition  is  a  general  term  for  either  a  theorem, 
or  a  problem. 

One  proposition  is  the  converse  of  another,  when  the  con- 
clusion of  the  first  is  made  the  supposition  in  the  second. 

26.  A  corollary  is  an  obvious  consequence,  resulting  from 
one  or  more  propositions. 

27.  A  scholium  is  a  remark  appended  to  a  proposition. 

28.  An  hypothesis  is  a  supposition  made  either  in  the  enun- 
ciation of  a  proposition,  or  in  the  course  of  a  demonstration. 


Axioms. 

1.  Things  which  are  equal  to  the  same  thing  are  equal  to 
each  other. 

2.  If  equals  are  added  to  equals,  the  wholes  are  equal. 

3.  If  equals  are  taken  from  equals,  the  remainders   are 
equal. 

4.  If  equals  are  added  to  unequals,  the  wholes  are  unequal. 

5.  If  equals  are  taken  from  unequals,  the  remainders  are 
unequal. 

6.  Things  which  are  doubles  of  the  same  thing  are  equal  to 
each  other. 

7.  Things  which  are  halves  of  the  same  thing  are  equal  to 
each  other. 

8.  Magnitudes  which   coincide  with  each  other,  that  is, 
which  exactly  fill  the  same  space,  are  equal. 

9.  The  whole  is  greater  than  any  of  its  parts. 

10.  The  whole  is  equal  to  the  sum  of  all  its  parts. 

11.  From  one  point  to  another  only  one  straight  line  can 
be  drawn. 


BOOK    1.  13 


12.  Twc  straight  lines,  which  intersect  one  another  can 
not  both  be  parallel  to  the  same  straight  line. 


Explanation  of  Signs. 

For  the  sake  of  brevity,  it  is  convenient  to  employ,  to  soma 
extent,  the  signs  of  Algebra  in  Geometry.  Those  chiefly  em 
ployed  are  the  following : 

The  sign  =  denotes  that  the  quantities  between  which  it 
stands  are  equal ;  thus,  the  expression  A=B  signifies  that  A 
is  equal  to  B. 

The  sign  +  is  called  plus,  and  indicates  addition ;  thus 
A-f  B  represents  the  sum  of  the  quantities  A  and  B. 

The  sign  —  is  called  minus,  and  indicates  subtraction  ;  thus, 
A — B  represents  what  remains  after  subtracting  B  from  A. 

The  sign  X  indicates  multiplication  ;  thus,  AxB  denotes 
the  product  of  A  by  B.  Instead  of  the  sign  X,  a  point  is 
sometimes  employed  ;  thus,  A .  B  is  the  same  as  A  X  B.  The 
same  product  is  also  sometimes  represented  without  any  in- 
termediate sign,  by  AB ;  but  this  expression  should  not  be 
employed  when  there  is  any  danger  of  confounding  it  with 
the  line  AB. 

A  parenthesis  (  )  indicates  tHat  several  quantities  are  to 
be  subjected  to  the  same  operation ;  thus,  the  expression 
Ax(B+C — D)  represents  the  product  of  A  by  the  quantity 
B+C  — D. 

A 

The  expression  -^  indicates  the  quotient  arising  from  divi 

ding  A  by  B. 

A  number  placed  before  a  line  or  a  quantity  is  to  be  re 
garded  as  a  multiplier  of  that  line  or  quantity ;  thus,  SAB  de 
notes  that  the  line  AB  is  taken  three  times ;  $A  denotes  the 
half  of  A. 

The  square  of  the  line  AB  is  denoted  by  AB2 ;  its  cube  b) 
AB3. 

The  sign  ^/  indicates  a  root  to  be  extracted  ;  thus,  ^/2  de- 
notes the  square  root  of  2  •  \/A  X  B  denotes  the  square  root 
of  the  product  of  A  and  B. 

N.B. — The  first  six  books  treat  only  of  plane  jigures,  or  fe 
ures  drawn  on  a  plane  surface. 


14  GEOMETRY. 

PROPOSITION    I.       THEOREM. 

All  1  ight  angles  are  equal  to  each  nther. 

Lei  the  straight  line  CD  D    -K  IL 

be  perpendicular  to  AB,  and 
GH  to  EF ;  then,  by  defini- 
tion 10,  each  of  the  angles 

ACD,  BCD,  EGH,  FGH,  will    

be  a  right  angle  ;  and  it  is  to 

be  proved  that  the  angle  ACD  is  equal  to  the  angle  EGH. 

Take  the  four  straight  lines  AC,  CB,  EG,  GF,  all  equal  to 
each  other ;  then  will  the  line  AB  be  equal  to  the  line  EF 
(Axiom  2).  Let  the  line  EF  be  applied  to  the  line  AB, 
so  that  the  point  E  may  be  on  A,  and  the  point  F  on  B ; 
then  will  the  lines  EF,  AB  coincide  throughout ;  for  other- 
wise two  different  straight  lines  might  be  drawn  from  one 
point  to  another,  which  is  impossible  (Axiom  11).  More- 
over, since  the  line  EG  is  equal  to  the  line  AC,  the  point  G 
will  fall  on  the  point  C ;  and  the  line  EG,  coinciding  with 
AC,  the  line  GH  will  coincide  with  CD.  For,  if  it  could 
have  any  other  position,  as  CK,  then,  because  the  angle  EGH 
is  equal  to  FGH  (Def.  10),  the  angle  ACK  must  be  equal  to 
BCK,  and  therefore  the  angle  ACD  is  less  than  BCK.  Bui 
BCK  is  less  than  BCD  (Axiom  9) ;  much  more,  then,  is  ACD 
less  than  BCD,  which  is  impossible,  because  the  angle  ACD 
is  equal  to  the  angle  BCD  (Def.  10) ;  therefore,  GH  can  not 
but  coincide  with  CD,  and  the  angle  EGH  coincides  with 
the  angle  ACD,  and  is  equal  to  it  (Axiom  8).  Therefore,  all 
riejht  angles  are  eoual  to  each  other. 

PROPOSITION    II.       THEOREM, 

The  angles  which  one  straight  line  makes  w>tk  another,  up  j* 
one  side  of  it,  are  either  two  right  angles,  or  are  together  eqiu< 
to  two  right  angles. 

Let  the  straight  line  AB  make  with  CD, 
upon  one  side  of  it,  the  angles  ABC,  ABD ; 
these  are  either  two  right  angles,  or  are  to- 
gether equal  to  two  right  angles. 

For  if  the  angle  ABC  is  equal  to  ABD, 

each  of  them  is  a  right  angle  (Def  10)  ;  but 


BOOK    1. 

.1  not,  suppose  the  line  BE  to  be  drawn  from  E| 

the  point  B,  perpendicular  to  CD ;  then  will 
each  of  the  angles  CBE,  DBE  be  a  right 
angle.  Now  the  angle  CBA  is  equal  to  the 
sum  of  the  two  angles  CBE,  EBA.  To 
each  of  these  equals  add  the  angle  ABD;  c 
then  the  sum  of  the  two  angles  CBA,  ABD  will  be  equal  to 
the  sum  of  the  three  angles  CBE,  EBA,  ABD  (Axiom  2). 
Again,  the  angle  DBE  is  equal  to  the  sum  of  the  two  angles 
DBA,  ABE.  Add  to  each  of  these  equals  the  angle  EBC; 
then  will  the  sum  of  the  two  angles  DBE,  EBC  be  equal  to 
the  sum  of  the  three  angles  DBA,  ABE,  EBC.  Now  things 
that  are  equal  to  the  same  thing  are  equal  to  each  other 
(Axiom  1) ;  therefore,  the  sum  of  the  angles  CBA,  ABD  is 
equal  to  the  sum  of  the  angles  CBE,  EBD.  But  CBE,  EBD 
are  two  right  angles ;  therefore  ABC,  ABD  are  together 
equal  to  two  right  angles.  Therefore,  the  angles  which  one 
straight  line,  &c. 

Corollary  1.  If  one  of  the  angles  ABC.  ABD  is  a  right 
angle,  the  other  is  also  a  right  angle. 

Cor.  2.  If  the  line  DE  is  perpendicular  to  D 

AB,  conversely,  AB  will  be  perpendicular  to 
DE.  _ 

For,  because  DE  is  perpendicular  to  AB,  A 
the  angle  DCA  must  be  equal  to  its  adjacent 
angle  DCB  (Def.  10),  and  each  of  them  must 
be  a  right  angle.  But  since  ACD  is  a  right  angle,  its  adja- 
cent angle,  ACE,  must  also  be  a  right  angle  (Cor.  1).  Hence 
the  angle  ACE  is  equal  to  the  angle  ACD  (Prop.  I.),  and  AB 
is  perpendicular  to  DE. 

Cor.  3.  The  sum  of  all  the  angles  BAG,  Dy 

CAD,  DAE,  EAF,  formed  on  the  same  /     /E 

side  of  the  line  BF,  is  equal  to  two  right 
angles;  for  their  sum  is  equal  to  that  of 
the  two  adjacent  angles  BAD,  DAF.  B" 


PROPOSITION  in.     THEOREM  (Converse  of  Prop.  II.). 

Iff  at  a  point  in  a  straight  line,  two  other  straight  lines,  upon 
tne  opposite  sides  of  it,  make  the  adjacent  angles  together  equal 
to  two  right  angles,  these  two  straight  lines  are  in  one  and  th? 
same  straight  line. 

At  the  point  B,  in  the  straight  line  AB,  let  the  two  straight 
linos  BC,  BD,  upon  the  opposite  sides  of  AB,  make  the  adja- 
cent angles,  ABC,  ABD,  together  equal  to  two  righ  angles- 


16  GEOMETRY 

then  will  BD  be  in  the  same  straight  line 
with  CB. 

For,  if  BD  is  not  in  the  same  straight 
line  with  CB,  let  BE  be  in  the  same 

straight  line  with  it;  then,  because  the 

straight  line  CBE  is  met  by  the  straight  c 
line  AB,  the  angles  ABC,  ABE  are  together  equal  to  two 
right  angles  (Prop.  II.).  But,  by  hypothesis,  the  angles  ABC, 
ABD  are  together  equal  to  two  right  angles  ;  therefore,  the 
sum  of  the  angles  ABC,  ABE  is  equal  to  the  sum  of  the  an- 
gles ABC,  ABD.  Take  away  the  common  angle  ABC,  and 
the  remaining  angle  ABE,  is  equal  (Axiom  3)  to  the  remain- 
ing angle  ABD,  the  less  to  the  greater,  which  is  impossible. 
Hence  BE  is  not  in  the  same  straight  line  with  BC ;  and  in 
like  manner,  it  may  be  proved  that  no  other  can  be  in  the  same 
straight  line  with  it  but  BD.  Therefore,  if  at  a  point,  &c. 


PROPOSITION   IV.       THEOREM. 

Two  straight  lines,  which  have  two  points  common,  coincide 
with  each  other  throughout  their  whole  extent,  and  form  but  one 
and  the  same  straight  line. 

Let  there  be  two  straight  lines,  having  p. 

the  points  A  and  B  in  common ;  these 
h'nes  will  coincide  throughout  their  whole 
extent.  E 

It  is  plain  that  the  two  lines  must  co-  - — = — g —  ^~j\ 
incide  between  A  and  B,  for  otherwise 
there  would  be  two  straight  lines' between  A  and  B,  which 
is  impossible  (Axiom  11).  Suppose,  however,  that,  on  being 
produced,  these  lines  begin  to  diverge  at  the  point  C,  one 
taking  the  direction  CD,  anc.  the  other  CE.  From  the  point 
C  draw  the  line  CF  at  rignt  angles  with  AC ;  then,  since 
A  CD  is  a  straight  line,  the  angle  FCD  is  a  right  angle  (Prop. 
II ,  Cor.  1)  ;  and  since  ACE  is  a  straight  line,  the  angle  FCE 
is  also  a  right  angle  ;  therefore  (Prop.  I.),  the  angle  FCB 
is  equal  to  the  angle  FCD,  the  less  to  the  greater,  which  u 
absurd.  Therefore,  two  straight  lines  which  have,  &c. 


PROPOSITION    V.       THEOREM. 


If  two  straight  lines  cut  one  another,  the  vertical  or  opposib*. 
angles  are  equal. 

I  et  the  two  straigh.  lines.  AB,  CD.  cut  one  another  in  the 


BOOK    I. 


point  E  ;  then  will  the  angle  AEG  be  equal 
to  the  angle  BED,  and  the  angle  AED  to 
the  angle  CEB.  __  _ 

For  the  angles  AEC,  AED,  which  the  A  ~~^\  B 
straight  line  AE  makes  with  the  straight  ;NV 

line  CD,  are  together  equal  to  two  right 
angles  (Prop.  II.)  ;  and  the  angles  AED,  DEB,  which  the 
straight  line  DE  makes  with  the  straight  line  AB,  are  also 
together  equal  to  two  right  angles  ;  therefore,  the  sum  of  the 
r.wo  angles  AEC,  AED  is  equal  to  the  sum  of  the  two  angles 
,YED,  DEB.  Take  away  the  common  angle  AED,  and  the 
-•emaining  angle,  AEC,  is  equal  to  the  remaining  angle  DEB 
v  Axiom  3).  In  the  same  manner,  it  may  be  proved  that  the 
angle  AED  is  equal  to  the  angle  CEB.  Therefore,  if  two 
straight  lines,  &c. 

Cor.  1.  Hence,  if  two  straight  lines  cut  one  another,  the 
four  angles  formed  at  the  point  of  intersection,  are  together 
equal  to  four  right  angles. 

Cor.  2.  Hence,  all  the  angles  made  by  any  number  of 
straight  lines  meeting  in  one  point,  are  together  equal  to  four 
right  angles. 


PROPOSITION    VI.       THEOREM. 

If  two  triangles  have  two  sides,  and  the  included  angle  of  the 
vne,  equal  to  two  sides  and  the  included  angle  of  the  other,  each 
to  each,  the  two  triangles  will  be  equal,  their  third  sides  will  be. 
equal,  and  their  other  angles  will  be  equal,  each  to  each. 

Let  ABC,  DEF  be  two  triangles, 
having  the  side  AB  equal  to  DE, 
and  AC  to  DF,  and  also  the  angle 
A  equal  to  the  angle  D;  then  will 
:he  triangle  ABC  be  equal  to  the 
triangle  DEF. 

For,  if  the  triangle  ABC  is  ap- 
plied  to  the  triangle  DEF,  so  that  the  point  A  may  be  on  D, 
and  the  straight  line  AB  upon  DE,  the  point  B  will  coincide 
with  the  point  E,  because  AB  is  equal  to  DE ;  and  AB,  coin- 
ciding with  DE,  AC  will  coincide  with  DF,  because  the  an- 
gle A  is  equal  to  the  angle  D.  Hence,  also,  the  point  C  will 
coincide  with  the  point  F,  because  AC  is  equal  to  DF.  But 
the  point  B  coincides  with  the  point  E  ;  therefore  the  base 
BC  will  coincide  with  the  base  EF  (Axiom  11),  and  will  be 
equal  to  it.  Hence,  also,  the  whole  triangle  ABC  will  coin 
cide  with  the  whole  triangle  DEF,  and  will  be  equal  to  it 

B 


18  GEOMETRY. 

and  the  remaining  angles  of  the  one,  will  coincide  with  the 
remaining  angles  of  the  other,  and  be  equal  to  them,  viz. :  the 
angle  ABC  to  the  angle  DEF,  and  the  angle  ACB  to  the  an- 
gle DFE.  Therefore,  if  two  triangles,  &c. 


PROPOSITION    VII.       THEOREM. 

If  two  triangles  have  two  angles,  and  the  included  side  of  th* 
one,  equal  to  two  angles  and  the  included  side  of  the  other,  each 
to  each,  the  two  triangles  will  be  equal,  the  other  sides  will  be 
equal,  each  to  each,  and  the  third  angle  of  the  one  to  the  third 
angle  of  the,  other. 

Let  ABC,  DEF  be  two 
triangles  having  the  angle 
B  equal  to  E,  the  angle  C 
equal  to  F,  and  the  inclu- 
ded sides  BC,  EF  equal  to  

each  other;    then  will  the  B  CE 

triangle  ABC  be  equal  to  the  triangle  DEF. 

For,  if  the  triangle  ABC  is  applied  to*  the  triangle  DEF,  so 
that  the  point  B  may  be  on  E,  and  the  straight  line  BC  upon 
EF,  the  point  C  will  coincide  with  the  point  F,  because  BC 
is  equal  to  EF.  Also,  since  the  angle  B  is  equal  to  the  an- 
gle E,  the  side  BA  will  take  the  direction  ED,  and  therefore 
the  point  A  will  be  found  somewhere  in  the  line  DE.  And 
because  the  angle  C  is  equal  to  the  angle  F,  the  line  CA  will 
take  the  direction  FD,  and  the  point  A  will  be  found  some- 
where in  the  line  DF ;  therefore,  the  point  A,  being  found  at 
the  same  time  in  the  two  straight  lines  DE,  DF,  must  fall  ai 
their  intersection,  D.  Hence  the  two  triangles  ABC,  DEF 
coincide  throughout,  and  are  equal  to  each  other ;  also,  the 
two  sides  AB,  AC  are  equal  to  the  two  sides  DE,  DF,  each 
to  each,  and  the  angle  A  to  the  angle  D.  Therefore,  if  two 
triangles,  &c. 


PROPOSITION    VIII.       THEOREM. 

Any  side  of  a  triangle  is  less  than  the  sum  of  the  other  two 

Let  ABC  be  a  triangle ;  any  one  of  its 
sides  is  less  than  the  sum  of  the  other  two, 
viz. :  the  side  AB  is  less  than  the  sum  of  AC 
and  BC  ;  BC  is  less  than  the  sum  of  AB  and 
AC  ;  and  AC  is  less  than  the  sum  of  AB  B 
and  BC. 


BOOK    I.  19 


For  the  straight  line  AB  is  the  shortest  path  between  tne 
points  A,  and  B  (Def.  3) ;  hence  AB  is  less  than  the  sum  of 
AC  and  BC.  For  the  same  reason,  EC  is  less  than  the  sum 
of  AB  and  AC ;  and  AC  less  than  the  sum  of  AB  and  BC 
Therefore,  any  two  sides,  &c. 


PROPOSITION   IX.       THEOREM. 

lff  from  a  point  withir*  a  triangle,  two  straight  lines  are 
drawn  to  the  extremities  of  either  side,  their  sum  will  be  less 
han  the  sum  of  the  other  two  sides  of  the  triangle. 

Let  the  two  straight  lines  BD,  CD  be 
drawn  from  D,  a  point  within  the  triangle 
ABC,  to  the  extremities  of  the  side  BC ; 
then  will  the  sum  of  BD  and  DC  be  less 
than  the  sum  of  BA,  AC,  the  other  two 
sides  of  the  triangle.  

Produce  BD  until  it  meets  the  side  AC  B  C 

in  E  ;  and,  because  one  side  of  a  triangle  is  less  than  the  sum 
of  the  other  two  (Prop.  VIII.),  the  side  CD  of  the  triangle 
CDE  is  less  than  the  sum  of  CE  and  ED.  To  each  of  these 
add  DB;  then  will  the  sum  of  CD  and  BD  be  less  than  the 
sum  of  CE  and  EB.  Again,  because  the  side  BE  of  the  tri- 
angle BAE  is  less  than  the  sum  of  BA  and  AE,  if  EC  be  add- 
ed to  each,  the  sum  of  BE  and  EC  will  be  less  than  the  sum 
of  BA  and  AC.  But  it  has  been  proved  that  the  sum  of  BD 
and  DC  is  less  than  the  sum  of  BE  and  EC  ;  much  more,  then, 
is  the  sum  of  BD  and  DC  less  than  the  suni  of  BA  and  AC, 
Therefore,  if  from  a  point,  &c. 


PROPOSITION    X.       THEOREM. 

The  angles  at  the  base  of  an  isosceles  triangle  are  equal  to 
one  another. 

Let  ABC  be  an  isosceles  triangle,  of  which 
the  side  AB  is  equal  to  AC ;  then  will  the  angle 
B  be  equal  to  the  angle  C. 

For,  conceive  the  angle  BAG  to  be  bisected 
by  the  straight  line  AD  ;  then,  in  the  two  trian- 
gles ABD,  ACD,  two  sides  AB,  AD,  and  the  in- 
cluded angle  in  the  one,  are  equal  to  the  two  B  D  C 
sides  AC,  AD,  and  the  included  angle  in  the  other ;  there- 
fore (Prop.  VI.),  the  angle  B  is  equal  to  the  angle  C.  There- 
tore,  the  angles  at  the  base,  &c. 


20  GEOMETRY. 

Cor.  1.  Hence,  also,  the  line  BD  is  equal  to  DC,  and  the 
angle  ADB  equal,  to  ADC ;  consequently,  each  of  these  an- 
gles is  a  right  angle  (Def.  10).  Therefore,  the  line  bisecting 
the  vertical  angle  of  an  isosceles  triangle  bisects  the  base  ai 
right  angles  ;  and,  conversely,  the  line  bisecting  the  base  of  an 
isosceles  triangle  2t  right  angles  bisects  also  the  vertical  angle. 

Cor.  2.  Every  equilateral  triangle  is  also  equiangular. 

Scholium.     Any  side  of  a  triangle  may  be  considered  as 
its  base,  and  the  opposite  angle  as  its  vertex ;  but  in  an  isos 
celes  triangle,  that  side  is  usually  regarded  as  the  base,which 
is  not  equal  to  either  of  the  others. 


PROPOSITION  xi.     THEOREM  (Converse  of  Prop.  X.). 

If  two  angles  of  a  triangle  are  equal  to  one  another,  the  op- 
posite sides  are  also  equal. 

Let  ABC  be  a  triangle  having  the  angle 
ABC  equal  to  the  angle  ACB ;  then  will  the 
side  AB  be  equal  to  the  side  AC. 

For  if  AB  is  not  equal  to  AC,  one  of  them 
must  be  greater  than  the  other.  Let  AB  be 
the  greater,  and  from  it  cut  off  DB  equal  to  AC 
the  less,  and  join  CD.  Then,  because  in  the  tri-  _ 
angles  DBC,  ACB,  DB  is  equal  to  AC,  and  BC  J 
is  common  to  both  triangles,  also,  by  supposition,  the  angle 
DBC  is  equal  to  the  angle  ACB ;  therefore,  the  triangle  DBC 
is  equal  to  the  triangle  ACB  (Prop.  VI.),  the  less  to  the  great- 
er, which  is  absurd.  Hence  AB  is  not  unequal  to  AC,  that 
>-s,  it  is  equal  to  it.  Therefore,  if  two  angles,  &c. 

Cor.  Hence,  every  equiangular  triangle  is  also  equilateral. 


PROPOSITION   XII.       THEOREM. 

The  greater  side  of  every  triangle  is  opposite  to  the  greater 
angle ;  and,  conversely,  the  greater  angle  is  opposite  to  thti 
greater  side. 

Let  ABC  be  a  triangle,  having  the  angle  ACB 
greater  than  the  angle  ABC  ;  then  will  the  side        /\D 
AB  be  greater  than  the  side  AC. 

Draw  the  straight  line  CD,  making  the  angle 
BCD  equal  to  B  ;  then,  in  the  triangle  CDB,  the 
side  CD  must  be  equal  to  DB  (Prop.  XL).  Add 
AD  to  each,  then  will  the  sum  of  AD  and  DC 


BOOK    1.  21 

be  equal  to  the  sum  of  AD  and  DB.  But  AC  is  less  tnan  the 
sum  of  AD  and  DO  (Prop.  VIII.) ;  it  is,  therefore,  less  than 
AB. 

Conversely,  if  the  side  AB  is  greater  than  the  side  AC,  then 
will  the  angle  ACB  be  greater  than  the  angle  ABC. 

For  if  ACB  is  not  greater  than  ABC,  it  must  be  either 
equal  to  it,  or  less.  It  is  not  equal,  because  then  the  side 
AB  would  be  equal  to  the  side  AC  (Prop.  XL),  which  is  con- 
trary to  the  supposition.  Neither  is  it  less,  because  then  the 
side  AB  would  be  less  than  the  side  AC,  according  to  the  for- 
mer part  of  this  proposition ;  hence  ACB  must  be  greater 
than  ABC.  Therefore,  the  greater  side,  &c. 


PROPOSITION  XIII.       THEOREM. 

If  two  triangles  have  two  sides  of  the  one  equal  to  two  siaes 
of  the  other,  each  to  each,  but  the  included  angles  unequal,  the 
base  of  that  which  has  the  greater  angle , will  be  greater  thar 
the  base  of  the  other. 

Let  ABC,  DEF  be  two  trian- 
gles, having  two  sides  of  the  one 
equal  to  two  sides  of  the  other, 
'- iz. :  AB  equal  to  DE,  and  AC  to 
DF,  but  the  angle  BAG  greater 
than  the  angle  EDF;  then  will 
the  base  BC  be  greater  than  the 
base  EF. 

Of  the  two  sides  DE,  DF,  let  DE  be  the  side  which  is  not 
greater  than  the  other ;  and  at  the  point  D,  in  the  straight 
line  DE,  make  the  angle  EDG  equal  to  BAG ;  make  DG 
equal  to  AC  or  DF,  and  join  EG,  GF. 

Because,  in  the  triangles  ABC,  DEG,  AB  is  equal  to  DE, 
and  AC  to  DG ;  also,  the  angle  BAG  is  equal  to  the  angle 
EDG ;  therefore,  the  base  BC  is  equal  to  the  base  EG  (Prop. 
VI.).  Also,  because  DG  is  equal  to  DF,  the  angle  DFG  is 
equal  to  the  angle  DGF  (Prop.  X.).  But  the  angle  DGF  is 
greater  than  the  angle  EGF ;  therefore  the  angle  DFG  is 
greater  than  EGF  ;  and  much  more  is  the  angle  EFG  greater 
than  the  angle  EGF.  Now,  in  the  triangle  EFG,  because 
the  angle  EFG  is  greater  than  EGF,  and  because  the  great 
er  side  is  opposite  the  greater  angle  (Prop.  XIL),  the  side 
EG  is  greater  than  the  side  EF.  But  EG  has  been  proved 
equal  to  BC  ;  and  hence  BC  is  greater  than  EF.  Therefore, 
3  two  triangles,  &c. 


22 


GEOMETRY. 


/ROPUS:TION  xiv.     THEOREM  (Convetse  of  Prop    XT//.). 

If  two  triangles  have  two  sides  of  the  one  equal  #.;  two  sides 
of  the  other,  each  to  each,  but  the  bases  unequal,  the  angle  con- 
tained by  the  sides  of  that  which  has  the  greater  base,  will  be 
greater  than  the  angle  contained  by  the  sides  of  the  other. 

Let  ABC,  DEF  be  two  triangles 
having  two  sides  of  the  one  equal  to 
two  sides  of  the  other,  viz. :  AB  equal 
to  DE,  and  AC  to  DF,  but  the  base 
BC  greater  than  the  base  EF;  then 
will  the  angle  BAG  be  greater  than 
the  angle  EDF. 

For  if  it  is  not  greater,  it  must  be 
either  equal  to  it,  or  less.  But  the  angle  BAG  is  not  equal 
to  the  angle  EDF,  because  then  the  base  BC  would  be  equal 
to  the  base  EF  (Prop.  VI.),  which  is  contrary  to  the  suppo- 
sition. Neither  is  it  less,  because  then  the  base  BC  would  be 
less  than  the  base  EF  (Prop.  XIII.) ,  which  is  also  contrary 
to  the  supposition ;  therefore,  the  angle  BAG  is  not  less  than 
the  angle  EDF,  and  it  has  been  proved  that  it  is  not  equal 
to  it ;  hence  the  angle  BAG  must  be  greater  than  the  angle 
EDF.  Therefore,  if  two  triangles,  &c. 


PROPOSITION   XV.       THEOREM. 

If  two  triangles  have  the  three  sides  of  the  one  equal  to  tfie 
three  sides  of  the  other,  each  to  each,  the  three  angles  will  also 
be  equal,  each  to  each,  and  the  triangles  themselves  will  le 
equal 

Let  ABC,  DEF  be  two  trian- 
gles having  the  three  sides  of  the 
one  equal  to  the  three  sides  of  the 
other,  viz. :  AB  equal  to  DE,  BC 
to  EF,  and  AC  to  DF  ;  then  will 
the  three  angles  also  be  equal, 
viz. :  the  angle  A  to  the  angle  D, 
the  angle  B  to  the  angle  E,  and  the  angle  C  to  the  angle  F. 

For  if  the  angle  A  is  not  equal  to  the  angle  D,  it  must  be 
either  greater  or  less.  It  is  not  greater,  because  then  tie 
base  BC  would  be  greater  than  the  base  EF  (Prop.  XIII ) 
which  is  contrary  to  the  hypothesis;  neithe:  IE  it  less,  be 


BOOK    I.  23 

cause  then  the  base  BC  would  be  less  than  the  base  El 
(Prop.  XIII.) ,  which  is  also  contrary  to  the  hypothesis. 
Therefore,  the  angle  A  must  be  equal  to  .the  angle  D.  In 
the  same  manner,  it  may  be  proved  that  the  angle  B  is  equal 
to  the  angle  E,  and  the  angle  C  to  the  angle  F ;  hence  the 
two  triangles  are  equal.  Therefore,  if  two  triangles,  &c. 

Scholium.  In  equal  triangles,  the  equal  angles  are  oppo 
site  to  the  equal  sides ;  thus,  the  equal  angles  A  and  D  are 
opposite  to  the  equal  sides  BC,  EF. 


PROPOSITION    XVI.       THEOREM. 

From  a  point  without  a  straight  line*  only  one  perpendicular 
can  be  drawn  to  that  line. 

Let  A  be  the  given  point,  and  DE  the  _A_ 

given  straight  line ;  from  the  point  A  only 
one  perpendicular  can  be  drawn  to  DE. 

For,  if  possible,  let  there  be  drawn  two 
perpendiculars  AB,  AC.  Produce  the  line 
AB  to  F,  making  BF  equal  to  AB,  and  join 
CF.  Then,  in  the  triangles  ABC,  FBC,  be- 
cause AB  is  equal  to  BF,  BC  is  common  to 
both  triangles,  and  the  angle  ABC  is  equal  to  the  angle  FBC, 
being  both  right  angles  (Prop.  II.,  Cor.  1);  therefore,  two 
sides  and  the  included  angle  of  one  triangle,  are  equal  to  two 
sides  and  the  included  angle  of  the  other  triangle ;  hence  the 
angle  ACB  is  equal  to  the  angle  FCB  (Prop.  VI.).  But, 
since  the  angle  ACB  is,  by  supposition,  a  right  angle,  FCB 
must  also  be  a  right  angle;  and  the  two  adjacent  angles 
BCA,  BCF,  being  together  equal  to  two  right  angles,  the  two 
straight  lines  AC,  CF  must  form  one  and  the  same  straight 
line  (Prop.  III.) ;  that  is,  between  the  two  points  A  and  F, 
two  straight  lines,  ABF,  ACF,  may  be  drawn,  which  is  im- 
possible (Axiom  11)  ;  hence  AB  and  AC  can  not  both  be  per 
pendicular  to  DE.  Therefore,  from  a  point,  &c. 

Cor.  From  the  same  point,  C,  in  the 
line  AB,  more  than  one  perpendicular  to 
this  line  can  not  be  drawn.  For,  if  possi- 
ble, let  CD  and  CE  be  two  perpendicu- 
lars ;  then,  because  CD  is  perpendicular 

to  AB,  the  angle  DCA  is  a  right  angle ;  A. ^ B 

and,  because  CE  is  perpendicular  to  AB, 

the  angle  EGA  is  also  a  right  angle.     Hence,  the  angle  ACD 

is  equal  to  the  angle  ACE  ^Prop.  I.),  the  less  to  the  greater 


\ 


GEOMETRY. 


which  is  absurd  ;  therefore,  CD  and  CE  can  not  both  be  pe 
pendicular  to  AB  from  the  same  point  C. 


PROPOSITION    XVII.       THE  DREM. 


Ifijrom  a  point  without  a  straight  line,  a  perpendicular  be 
drawn  to  this  line,  and  oblique  lines  be  drawn  to  different 
points  : 

1st.  TJie  perpendicular  will  be  shorter  than  any  oblique  line 

2d.  Two  oblique  lines,  which  meet  the  proposed  line  at  equa» 
distances  from  the  perpendicular,  will  be  equal. 

3d.  Of  any  two  oblique  lines,  that  which  is  further  from  tht 
perpendicular  will  be  the  longer. 

•Let  DE  be  the  given  straight  line,  and 
A  any  point  without  it.  Draw  AB  per- 
pendicular to  DE ;  draw,  also,  the  ob- 
lique lines  AC,  AD,  AE.  Produce  the 
line  AB  to  F,  making  BF  equal  to  AB, 
ind  join  CF,  DF. 

First.  Because,  in  the  triangles  ABC, 
FBC,  AB  is  equal  to  BF,  BC  is  common 
to  the  two  triangles,  and  the  angle  ABC  is  equal  to  the  angle 
FBC,  being  both  right  angles  (Prop.  II.,  Cor.  1)  ;  therefore, 
two  sides  and  the  included  angle  of  one  triangle,are  equal  to 
two  sides  and  the  included  angle  of  the  other  triangle  ;  hence 
the  side  CF  is  equal  to  the  side  CA  (Prop.  VI.).  But  the 
straight  line  ABF  is  shorter  than  the  broken  line  ACF  (Prop. 
VIII.) ;  hence  AB,  the  half  of  ABF,  is  shorter  than  AC,  the 
half  of  ACF.  Therefore,  the  perpendicular  AB  is  shorter 
than  any  oblique  line,  AC. 

Secondly.  Let  AC  and  AE  be  two  oblique  lines  which 
meet  the  line  DE  at  equal  distances  from  the  perpendicular ; 
they  will  be  equal  to  each  other.  For,  in  the  triangles  ABC, 
ABE,  BC  is  equal  to  BE,  AB  is  common  to  the  two  triangles, 
and  the  angle  ABC  is  equal  to  the  angle  ABE,  being  both 
right  angles  (Prop.  I.) ;  therefore,  two  sides  and  the  included 
angle  of  one  triangle  are  equal  to  two  sides  and  the  included 
angle  of  the  other ;  hence  the  side  AC  is  equal  to  the  side 
AE  (Prop.  VI.).  Wherefore,  two  oblique  lines,  equally  dis- 
tant from  the  perpendicular,  are  equal. 

Thirdly.  Let  AC,  AD  be  two  oblique  lines,  of  which  AD 
is  further  from  the  perpendicular  than  AC ;  then  will  AD  be 
longer  than  AC.  For  it  has  already  been  proved  that  AC  is 
equal  to  CF  ;  and  in  the  same  manner  it  may  be  proved  that 
AD  is  equal  to  DF.  Now,  by  Prop.  IX.,  the  sum  of  the  two 


BOOK    I.  2ft 

Hnes  AC,  CF  is  less  than  Jie  sum  of  the  two  lines  AD,  DP, 
Therefore,  AC,  the  half  of  ACF,  is  less  than  AD,  the  half  of 
ADF ;  hence  the  oblique  line  which  is  furthest  from  the  per 
pendicular  is  the  longest.  Therefore,  if  from  a  point,  &c. 

Cor.  1.  The  perpendicular  measures  the  shortest  distance 
of  a  point  from  a  line,  because  it  is  shorter  than  any  oblique 
/me. 

Cor.  2.  It  is  impossible  to  draw  three  equal  straight  lines 
from  the  same  point  to  a  given  straight  line. 


PROPOSITION    XVIII.       THEOREM. 

If  through  the  middle  point  of  a  straight  line  a  perpendic- 
ular is  drawn  to  this  line  : 

1st.  Each  point  in  the  perpendicular  is  equally  distant  from 
the  two  extremities  of  the  line. 

2d.  Any  point  out  of  the  perpendicular  is  unequally  dis 
tantfrom  those  extremities. 

Let  the  straight  line  EF  be  drawn  perpen- 
licular  to  AB  through  its  middle  point,  C. 

First.  Every  point  of  EF  is  equally  dis- 
tant from  the  extremities  of  the  line  AB ;  for, 
since  AC  is  equal  to  CB,  the  two  oblique 
lines  AD,  DB  are  equally  distant  from  the 
perpendicular,  and  are,  there  fore,  equal  (Prop. 
XVII.).  So,  also,  the  two  oblique  lines  AE, 
EB  are  equal,  and  the  oblique  lines  AF,  FB 
are  equal ;  therefore,  every  point  of  the  per- 
pendicular is  equally  distant  from  the  extremities  A  and  B, 

Secondly.  Let  I  be  any  point  out  of  the  perpendicular. 
Draw  the  straight  lines  IA.  IB ;  one  of  these  lines  must  cut 
the  perpendicular  in  some  point,  as  D.  Join  DB ;  then,  by 
the  first  case,  AD  is  equal  to  DB.  To  each  of  these  equals 
add  ID,  then  will  IA  be  equal  to  the  sum  of  ID  and  DB. 
Now,  in  the  triangle  IDB,  IB  is  less  than  the  sum  of  ID  and 
DB  (Prop.  VIII.) ;  it  is,  therefore,  less  than  IA  ;  hence,  every 
point  out  of  the  perpendicular  is  unequally  distant  from  the 
extremities  A  and  B.  Therefore,  if  through  the  middle 
point,  &c. 

Cor.  If  a  straight  line  have  two  points,  each  of  which  is 
equally  distant  from  the  extremities  of  a  second  line,  it  will 
De  perpendicular  to  the  second  line  at  its  middle  point. 

B 


26 


GEOMETRY. 


PROPOSITION    XIX.       THEOREM. 

If  two  right-angled  triangles  have  the  hypoJienuse  and  e 
side  of  the  one,equal  to  the  hypothenuse  and  a  side  of  the  other 
each  to  each,  the  triangles  are  equal. 

Let  ABC,  DEF  be  two 
right-angled  triangles,  having 
the  hypothenuse  AC  and  the 
side  AB  of  the  one,  equal  to 
the  hypothenuse  DF  and  side 


DE  of  the  other;   then  will 

the  side  BC  be  equal  to  EF,  and  the  triangle  ABC  to  the  tri 

angle  DEF. 

For  if  BC  is  not  equal  to  EF,  one  of  them  must  be  greater 
than  the  other.  Let  BC  be  the  greater,  and  from  it  cut  off 
BG  equal  to  EF  the  less,  and  join  AG.  Then,  in  the  triangles 
ABG,  DEF,  because  AB  is  equal  to  DE,  BG  is  equal  to  EF, 
and  the  angle  B  equal  to  the  angle  E,  both  of  them  being 
right  angles,  the  two  triangles  are  equal  (Prop.  VI.),  and  AG 
is  equal  to  DF.  But,  by  hypothesis,  AC  is  equal  to  DF,  and 
therefore  AG  is  equal  to  AC.  Now  the  oblique  line  AC,  be 
ing  further  from  the  perpendicular  than  AG,  is  the  longei 
(Prop.  XVIL),  and  it  has  been  proved  to  be  equal,  which  is 
impossible.  Hence  BC  is  not  unequal  to  EF,  that  is,  it  is  equa. 
to  it ;  and  the  triangle  ABC  is  equal  to  the  triangle  DEF 
(Prop.  XV.")  Therefore,  if  two  right-angled  triangles,  &c 


PROPOSITION  XX.       THEOREM. 

Two  straight  lines  perpendicular  to  a  thi~d  line,  are  pa?  - 


Let  the  two  straight  lines 
AC,  BD  be  both  perpendicu- 
lar  to  AB  ;  then  is  AC  par- 
allel to  BD. 

For  if  these  lines  are  not 
parallel,  being  produced,  they 
must  meet  on  one  side  or  the  other  of  AB.     Let  them  be  pro 
duced,  and  meet  in  O  ;  then  there  will  be  two  perpendicu- 
lars, OA,  OB,  let  fall    from  the   same   point,  on  the  same 
straight  line,  which  is  impossible  (Prop.  XVI.).     Therefore 
two  straight  lines,  &<» 


BOOK   I.  27 


PROPOSITION    XXI.       THEOREM. 

If  a  straight  line,  meeting  two  other  straight  lines,  makes  t/ie 
interior  angles  on  the  same  side,  together  equal  to  two  right  an- 
gles, the  two  lines  are  parallel. 

Let  the  straight  line  AB,  which  A  E c 
meets  the  two  straight  lines  AC,  BD, 
make  the  interior  angles  on  the  same 
side,  BAG,  ABD,  together  equal  to  two 
right  angles ;  then  is  AC  parallel  to 
BD. 

From  G,  the  middle  point  of  the  line 

AB,  draw  EGF  perpendicular  to  AC  ;  it  will  also  be  perpen- 
dicular to  BD.  For  the  sum  of  the  angles  ABD  and  ABF  is 
equal  to  two  right  angles  (Prop.  II.) ;  and  by  hypothesis  the 
sum  of  the  angles  ABD  and  BAG  is  equal  to  two  right  an- 
gles. Therefore,  the  sum  of  ABD  and  ABF  is  equal  to  the 
sum  of  ABD  and  BAG.  Take  away  the  common  angle 
ABD,  and  the  remainder,  ABF,  is  equal  to  BAG ;  that  is 
GBF  is  equal  to  GAE. 

Again,  the  angle  BGF  is  equal  to  the  angle  AGE  (Prop 
V.)  ;  and,  by  construction,  BG  is  equal  to  GA  ;  hence  the  tri- 
angles BGF,  AGE  have  two  angles  and  the  included  side  of 
the  one,  equal  to  two  angles  and  the  included  side  of  the  oth- 
er ;  they  are,  therefore,  equal  (Prop.  VII.)  ;  and  the  angle 
BFG  is  equal  to  the  angle  AEG.  But  AEG  is,  by  construc- 
tion, a  right  angle,  whence  BFG  is  also  a  right  angle ;  that 
is,  the  two  straight  lines  EC,  FD  are  perpendicular  to  the 
same  straight  line,  and  are  consequently  parallel  (Prop. 
XX.).  Therefore,  if  a  straight  line,  &c. 

Scholium.     When  a  straight  line  ^ 

intersects  two  parallel  lines,  the  in-  / 

'terior  angles  on  the  same  side,  are  ~/ 

those  which  lie  within  the  parallels,  A -/- J3 

and  on  the  same  side  of  the  secant  / 

line,  as   AGH,  GHC  ;    also,  BGH, / ^ 

GHD.  /k 

Alternate   angles    lie   within   the  f. 

parallels,  on  different  sides  of  the  ^ 

secant  line,  and  are  not  adjacent  to  each  other,  as  AGH 
GHD  ;  also,  BGH,  GHC. 

Either  angle  without  the  parallels  being  called  ar.  exterior 
angle,  the  interior  and  opposite  angle  on  the  same  side,  lies 
within  the  parallels,  on  the  same  side  of  the  secant  line,  but 


GEOMETRY. 


aot  adjacent;  thus,  GHD  is  an  interior  angle  opposite  to  tha 
exterior  angle  EGB ;  so,  also,  with  the  angles  CHG,  AGE. 


PROPOSITION    XXII.       THEOREM. 

If  a  straight  line,  intersecting  two  other  straight  lines,  makes 
'he  alternate  angles  equal  to  each  other,  or  makes  an  exterior 
•ingle  equal  to  the  interior  and  opposite  upon  the  same  side  of 
the  secant  line,  these  two  lines  are  parallel. 

Let  the  straight  line  EF,  which  E 

intersects  the  two  straight  lines  AB,  / 

CD,  make  the  alternate  angles  AGH,  Q./ 

GHD  equal  to  each  other ;  then  AB  A ~J~~ 

is  parallel  to  CD.     For,  to  each  of  / 

the  equal  angles  AGH,  GHD,  add  ^ Z .D 

the  angle  HGB ;    then  the  sum  of  75 

AGH  and  HGB  will  be  equal  to  the  / 

vum  of  GHD  and  HGB.     But  AGH 

and  HGB  are  equal  to  two  right  angles  (Prop.  II.) ;  there- 
fore, GHD  and  HGB  are  equal  to  two  right  angles  ;  and 
hence  AB  is  parallel  to  CD  (Prop.  XXL). 

Again,  if  the  exterior  angle  EGB  is  equal  to  the  interior 
and  opposite  angle  GHD,  then  is  AB  parallel  to  CD.  For, 
the  angle  AGH  is  equal  to  the  angle  EGB  (Prop.  V.) ;  and, 
by  supposition,  EGB  is  equal  to  GHD ;  therefore  the  angle 
AGH  is  equal  to  the  angle  GHD,  and  they  are  alternate  an- 
gles ;  hence,  by  the  first  part  of  the  proposition,  AB  is  par- 
allel to  CD.  Therefore,  if  a  straight  line,  &c. 


PROPOSITION  XXIII.       THEOREM. 

(Converse  of  Propositions  XXI.  and  XXII.) 

If  a  straight  line  intersect  two  parallel  lines,  it  makes  trie 
alternate  angles  equal  to  each  other  ;  also,  any  exterior  angh 
equal  to  the  interior  and  opposite  on  the  same  side ;  and  the, 
two  intsrior  angles  on  the  same  side  together  equal  to  two  right 
angles. 

Let  the  straight  line  EF  intersect 
the  two  parallel  lines  AB,  CD ;  the 
alternate  angles  AGH,  GHD  are 
equal  to  each  other ;  the  exterior  an- 
gle EGB  is  equal  to  the  interior  and 

opposite    angle   on  the    same    side,    c 

GHD  ;  and  the  two  interior  angles  on 
the  same  side,  BGH,  GHD,  are  to- 
gether euuid  to  two  ripjhl  an^le- 


BOOK   I.  29 

For  if  AGH  is  not  equal  to  GHD,  through  G  draw  the 
line  KL,  making  the  angle  KGH  equal  to  GHD ;  then  KL 
must  be  parallel  to  CD  (Prop.  XXII.).  But,  by  supposition, 
AB  is  parallel  to  CD ;  therefore,  through  the  same  point,  G. 
two  straight  lines  have  been  drawn  parallel  to  CD,  which  is 
impossible  (Axiom  12).  Therefore,  the  angles  AGH,  GHD 
are  not  unequal,  that  is,  they  are  equal  to  each  other.  Now 
the  angle  AGH  is  equal  to  EGB  (Prop.  V.),  and  AGH  has 
been  proved  equal  to  GHD ;  therefore,  EGB  is  also  equa  to 
GHD.  Add  to  each  of  these  equals  the  angle  BGH ;  then 
will  the  sum  of  EGB,  BGH  be  equal  to  the  sum  of  BGH, 
GHD.  But  EGB,  BGH  are  equal  to  two  right  angles  (Prop. 
II.) ;  therefore,  also,  BGH.  GHD  are  equal  to  two  right  an 
gles.  Therefore,  if  a  straight  line,  &c 

Cor.  1.  If  a  straight  line  is  perpendicular  to  one  of  twc 
parallel  lines,  it  is  also  perpendicular  to  the  other. 

Cor.  2.  If  two  lines,  KL  and  CD,  make  with  EF  the  twc 
angles  KGH,  GHC  together  less  than  two  right  angles,  then 
will  KL  and  CD  meet,  if  sufficiently  produced. 

For  if  they  do  not  meet,  they  are  parallel  (Def.  12).  Bui 
they  are  not  parallel ;  for  then  the  angles  KGH,  GHC  would 
be  equal  to  two  right  angles. 


PROPOSITION    XXIV.       THEOREM. 

Straight  lines  which  are  parallel  to  the  same  line,  are  paral 
lei  to  each  other. 

Let  the  straight  lines  AB,  CD  be  i 

each  of  them  parallel  to  the  line  EF  ;  IE _! F 

then  will  AB  be  parallel  to  CD. 

For,   draw    any    straight    line,   as  C ^ D 

PQR,   perpendicular  to   EF.     Then, 

since  AB  is  parallel  to  EF,  PR,  which  A jp B 

is  perpendicular  to  EF,  will  also  be 

perpendicular  to  AB  (Prop.  XXIIL,  Cor.  1) ;  and  since  CD 
is  parallel  to  EF,  PR  will  also  be  perpendicular  to  CD. 
Hence,  AB  and  CD  are  both  perpendicular  to  the  same 
straight  line,  and  are  consequently  parallel  (Prop.  XX.). 
Therefore,  straight  lines  which  are  parallel,  &c. 


PROPOSITION    XXV.       THEOREM. 


Two  parallel  straight  lines  are  every  where  equally  distant 
from  each  other. 

Let  AB    CD  be  two  parallel  straight  lines.     From  anj 


30  GEOMETRj 

points,  E  ind  F,  in  one  of  them,  II 
draw  Lie  lines  EG,  FH  perpendic-  c  \"~ 
ular  to  AB ;  they  will  also  be  per- 
pendicular to  CD  (Prop.  XXIIL, 


'D 


Cor.  1).     Join  EH ;  then,  because  A      F  E    B 

EG  and  FH  are  perpendicular  to  the  same  straight  line  AB 
they  are  parallel  (Prop.  XX.)  ;  therefore,  the  alternate  an 
gles,  EHF,  HEG,  which  they  make  with  HE  are  equal 
(Prop.  XXIIL).  Again,  because  AB  is  parallel  to  CD,  the 
alternate  angles  GHE,  HEF  are  also  equal.  Therefore,  the 
triangles  HEF,  EHG  have  two  angles  of  the  one  equal  to 
two  angles  of  the  other,  each  to  each,  and  the  side  Eli  inclu 
ded  between  the  equal  angles,  common ;  hence  the  triangles 
are  equal  (Prop.  VII.) ;  and  the  line  EG,  which  measures  the 
distance  of  the  parallels  at  the  point  E,  is  equal  tu  the  line 
FH,  which  measures  the  distance  of  the  same  parallels  at  the 
point  F.  Therefore,  two  parallel  straight  lines,  &c. 


PROPOSITION    XXVI.        THEOREM. 

Two  angles  are  equal,  when  their  sides  are  parallel,  each  to 
each,  and  are  similarly  situated. 

Let  BAG,  DEF  be  two  angles,  having 
he  side  BA  parallel  to  DE,  and  AC  to 
EF;  the  two  angles  are  equal  to  each 
other. 

Produce  DE,  if  necessary,  until  it  meets 
AC  in  G.     Then,  because  EF  is  parallel 

to  GC,  the  angle  DEF  is  equal  to  DGC  ^ -^ =. 

(Prop.  XXIIL) ;  and  because  DG  is  par-  J 
allel  to  AB,  the  angle  DGC  is  equal  to  BAG ;  hence  the  an 
gle  DEF  is  equal  to  the  angle  BAG  (Axiom  1).     Therefore, 
two  angles,  &c. 

Scholium.  This  proposition  is  restricted  to  the  case  in 
which  the  sides  which  contain  the  angles  are  similarly  situ- 
ated ;  because,  if  we  produce  FE  to  H,  the  angle  DEH  has 
its  sides  parallel  to  those  of  the  angle  BAG  ;  but  the  two  an- 
gles are  not  equal. 


PROPOSITION   XXVII.       THEOREM. 

If  one  side  of  a  triangle  is  produced,  the  exterior  angle  i» 
equal  to  the  sum  of  the  two  interior  and  opposite  angles ;  and 
the  three  interior  angles  of  every  triangle  are  equal  to  two 
right  angles. 

Let  ABC  be  anv  plane  triano'e,  and  let  the  side  BC  be 


BOOK    1.  31 

produced  10  D  ;  then  will  the  ex 
tenor  angle  ACD  be  equal  to  the 
sum  of  the  two  interior  and  oppo- 
site angles  A  and  B ;  and  the  sum 
of  the  three  angles  ABC,  BCA, 
CAB  is  equal  to  two  right  angles.  B 

For,  conceive  CE  to  be  drawn  parallel  to  the  side  AB  of 
the  triangle  ;  then,  because  AB  is  parallel  to  CE,  and  AC 
meets  them,  the  alternate  angles  BAC,  ACE  are  equal  (Prop. 
XXIIL).  Again,  because  AB  is  parallel  to  CE,  and  BD 
meets  them,  the  exterior  angle  ECD  is  equal  to  the  interior 
and  opposite  angle  ABC.  But  the  angle  ACE  was  proved 
equal  to  BAC ;  therefore  the  whole  exterior  angle  ACD  is 
equal  to  the  two  interior  and  opposite  angles  CAB,  ABC 
(Axiom  2).  To  each  of  these  equals  add  the  angle  ACB ; 
then  will  the  sum  of  the  two  angles  ACD,  ACB  be  equal  to 
the  sum  of  the  three  angles  ABC,  BCA,  CAB.  But  the  an- 
gles ACD,  ACB  are  equal  to  two  right  angles  (Prop.  II.) ; 
hence,  also,  the  angles  ABC,  BCA,  CAB  are  together  equal 
to  two  right  angles.  Therefore,  if  one  side  of  a  triangle,  &c. 

Cor.  1.  If  the  sum  of  two  angles  of  a  triangle  is  given,  the 
third  may  be  found  by  subtracting  this  sum  from  two  right 
angles. 

Cor.  2.  If  two  angles  of  one  triangle  are  equal  to  two  an- 
gles of  another  triangle,  the  third  angles  are  equal,  and  the 
triangles  are  mutually  equiangular. 

Cor.  3.  A  triangle  can  have  but  one  right  angle ;  for  if 
there  were  two,  the  third  angle  would  be  nothing.  Still  less 
can  a  triangle  have  more  than  one  obtuse  angle. 

Cor.  4.  In  a  right-angled  triangle,  the  sum  of  the  two  acute 
angles  is  equal  to  one  right  angle. 

Cor.  5.  In  an  equilateral  triangle,  each  of  the  angles  is  one 
>hird  of  two  right  angles,  or  two  thirds  of  one  right  angle. 


PROPOSITION  XXVIII.       THEOREM. 

The  sum  of  all  the  interior  angles  of  a  polygon,  is  equal  to 
twice  as  many  right  angles,  wanting  four,  as  the  figurt  has 
sides 

Let  ABCDE  be  any  polygon ;  then  the  sum  of  all  its  inte- 
rior angles  A,  B,  C,  D,  E  is  equal  to  twice  as  many  right  an 
gles,  wanting  four,  as  the  figure  has  sides  (see  next  page). 

For,  from  any  point,  F,  within  it,  draw  lines  FA,  FB,  FC, 
&c  ,  to  all  the  angles  The  polygon  is  thus  divided  into  as 
many  tri  ingles  as  it  has  sides.  Now  the  sum  of  the  three* 


GEOMETRY 


angles  of  each  of  these  triangles,  is  equal 
to  two  right  angles  (Prop.  XXVII.) ; 
therefore  the  sum  of  the  angles  of  all  the 
triangles  is  equal  to  twice  as  many  right 
angles  as  the  polygon  has  sides.  But 
the  same  angles  are  equal  to  the  angles 
of  the  polygon,  together  with  the  angles 
at  the  point  F,  that  is,  together  with  four 
right  angles  (Prop.  V.,  Cor.  2).  Therefore  the  angles  of  the 
polygon  are  equal  to  twice  as  many  right  angles  as  the  fig- 
ure has  sides,  wanting  four  right  angles. 

Cor.  1.  The  sum  of  the  angles  of  a  quadrilateral  is  four 
right  angles ;  of  a  pentagon,  six  right  angles ;  of  a  hexagon, 
eight,  &c. 

Cor.  2.  All  the  exterior  angles  of  a  polygon  are  together 
equal  to  four  right  angles.  Because  every  interior  angle,  ABC, 
together  with  its  adjacent  exterior  an- 
gle, ABD,  is  equal  to  two  right  angles 
(Prop.  II.)  ;  therefore  The  sum  of  all  the 
interior  and  exterior  angles,  is  equal  to  .v 
twice  as  many  right  angles  as  the  poly- 
gon has  sides  ;  that  is,  they  are  equal  to 
all  the  interior  angles  of  the  polygon, 
together  with  four  right  angles.  Hence 
the  sum  of  the  exterior  angles  must  be 
equal  to  four  right  angles  (Axiom  3). 


PROPOSITION  XXIX.       THEOREM. 

The  opposite  sides  and  angles  of  a  parallelogram  are  equal 
tc  each  other. 

Let  ABDC  be  a  parallelogram ;  then  will  ^  B 

ts  opposite  sides  and  angles  be  equal   to    v — ^\ 

each  other.  \       /''"    \ 

Draw  the  diagonal  BC  ;  then,  because  AB       \X[ \ 

s  parallel  to  CD,  and  BC  meets  them,  the       C  D 

alternate  angles  ABC,  BCD  are  equal  to  each  other  (Prop. 
XXIIL).  Also,  because  AC  is  parallel  to  BD,  and  BC  meets 
them,  the  alternate  angles  BCA,  CBD  are  equal  to  each  oth- 
er. Hence  the  two  triangles  ABC,  BCD  have  two  angles, 
ABC,  BCA  of  the  one,  equal  to  two  angles,  BCD,  CBD,  of 
the  other,  each  to  each,  and  the  side  BC  included  between 
•.hese  equal  angles,  common  to  the  two  triangles ;  therefore 
their  other  sides  are  equal,  each  to  each,  and  the  third  angla 
of  the  one  to  the  third  angle  of  the  othei  (Prop.  VII.),  viz. 


BOOK    I.  SJ 

the  side  AB  to  the  side  CD,  and  AC  10  BD,  and  the  angle 
BAG  equal  to  the  angle  BDC.  Also,  because  the  angle  ABC 
is  equal  to  the  angle  BCD,  and  the  angle  CBD  to  the  angle 
BCA,  the  whole  angle  ABD  is  equal  to  the  whole  angle 
A.CD.  But  the  angle  BAG  has  been  proved  equal  to  the  an 
gle  BDC ;  therefore  the  opposite  sides  and  angles  of  a  par 
ailelogram  are  equal  to  each  other. 

Cor.  Two  parallels,  AB,  CD,  comprehended  between  two 
other  parallels,  AC,  BD,  are  equal ;  and  the  diagonal  BC  di 
vides  the  parallelogram  into  two  equal  triangles. 


PROPOSITION  xxx.     THEOREM  (Converse  of  Prop.  XXIX.) 

If  the  opposite  sides  of  a  quadrilateral  are  equal,  each  to 
each,  the  equal  sides  are  parallel,  and  the  figure  is  a  parallels 
gram. 

Let  ABDC  be  a  quadrilateral,  having  its  A  B 

opposite  sides  equal  to  each  other,  viz. :  the     V 
side  AB  equal  to  CD,  and  AC  to  BD ;  then 
will  the  equal  sides  be  parallel,  and  the  fig- 
ure will  be  a  parallelogram.  ^ 

Draw  the  diagonal  BC ;  then  the  triangles  ABC,  BCD 
have  all  the  sides  of  the  one  equal  to  the  corresponding  sides 
of  the  other,  each  to  each ;  therefore  the  angle  ABC  is  equal 
to  the  angle  BCD  (Prop.  XV.),  and,  consequently,  the  side 
AB  is  parallel  to  CD  (Prop.  XXII.).  For  a  like  reason,  AC 
is  parallel  to  BD ;  hence  the  quadrilateral  ABDC  is  a  par- 
allelogram. Therefore,  if  the  opposite  sides,  &c. 


PROPOSITION   XXXI.       THEOREM. 

If  two  opposite  sides  of  a  quadrilateral  are  equal  and  par 
ullel,  the  other  two  sides  are  equal  and  parallel,  and  the  figure 
i*  a  parallelogram. 

Let  ABDC  be  a  quadrilateral,  having  the  A B 

sides  AB,  CD  equal  and  parallel ;  then  will    V~~        /'\ 
the  sides  AC,  BD  be  also  equal  and  parallel,     \     /         \ 
and  the  figure  will  be  a  parallelogram.  \£ ^ 

Draw  the  diagonal   BC ;    then,  because 
AB  is  parallel  to  CD,  and  BC  meets  them,  the  alternate  an 
gles  ABC,  BCD  are  equal   (Prop.  XXIII).     Also,  because 
AB  is  equal  to  CD,  and  BC  is  common  to  the  two  triangles 
ABC  BCD,  the  two  triangles  ABC,  BCD  have  two  sides  and 


34  GEOMETR    . 

the  included  angle  of  the  one,  equal  to  two  sides  and  the  in- 
cluded angle  of  the  other  ;  therefore,  the  side  AC  is  equal 
to  BD  (Prop.  VL),  and  the  angle  ACB  to  the  angle  CBD 
And,,  because  the  straight  line  BC  meets  the  two  straight 
lines  AC,  BD,  making  the  alternate  angles  BCA,  CBD  equal 
to  each  other,  AC  is  parallel  to  BD  (Prop.  XXII.) ;  hence 
the  figure  ABDC  is  a  parallelogram.  Therefore,  if  two  op- 
posite sides,  &c. 


PROPOSITION   XXXII.       THEOREM. 

The  diagonals  of  every  parallelogram  bisect  each  other 

Let  ABDC  be  a  parallelogram  whose  di-  A 
agonals,  AD,  BC,  intersect  each  other  in  E  ;    v 
then  will  AE  be  equal  to  ED,  and  BE  to 
EC. 

Because  the  alternate  angles  ABE,  ECD       C  D 

are  equal  (Prop.  XXIII.),  and  also  the  alternate  angles  EAB, 
EDC,  the  triangles  ABE,  DCE  have  two  angles  in  the  one 
equal  to  two  angles  in  the  other,  each  to  each,  and  the  inclu- 
ded sides  AB,  CD  are  also  equal ;  hence  the  remaining  sides 
are  equal,  viz. :  AE  to  ED,  and  CE  to  EB.  Therefore,  the 
diagonals  of  every  parallelogram,  &c. 

Cor.  If  the  side  AB  is  equal  to  AC,  the  triangles  AEB, 
AEC  have  all  the  sides  of  the  one  equal  to  the  corresponding 
sides  of  the  other,  and  are  consequently  equal  ;  hence  the 
angle  AEB  will  equal  the  angle  AEC,  and  therefore  the  di 
vgunals  of  a  rhombus  bisect  each  other  at  right  angles 


BOOK  ii.  35 


BOOK  II. 

RATIO  AND  PROPORTION. 
On  the  Relation  of  Magnitudes  to  Numbers. 

THE  ratios  of  magnitudes  may  be  expressed  by  numbers 
either  exactly  or  approximately ;  and  in  the  latter  case,  the 
approximation  can  be  carried  to  any  required  degree  of  pre 
cision. 

Thus,  let  it  be  proposed  to  find  the  numerical  ratio  of  two 
straight  lines,  AB  and  CD. 

From  the  greater  line  AB,  cut    A  ,,    n  _ 

~,  °       ,  ,  ,  f*\T\         -"-  ill         VT    ±> 

off  a  part  equal  to  the  less,  CD,     ( , i      |   i 

as  many  times  as  possible ;  for 
example,  twice,  with  a  remain-    ^ 

der  EB.     From  CD,  cut  off  a     ' ' ' 

part  equal  to  the  remainder  EB  as  often  as  possible ;  for  ex 
ample,  once,  with  a  remainder  FD.  From  the  first  remain- 
der, BE,  cut  off  a  part  equal  to  FD  as  often  as  possible ;  for 
example,  once,  with  a  remainder  GB.  From  the  second  re- 
mainder, FD,  cut  off  a  part  equal  to  the  third,  GB,  as  many 
times  as  possible.  Continue  this  process  until  a  remainder  is 
found  which  is  contained  an  exact  number  of  times  in  the 
preceding  one.  This  last  remainder  will  be  the  common 
measure  of  the  proposed  lines ;  and  regarding  it  as  the  meas- 
uring unit,  we  may  easily  find  the  values  of  the  preceding 
remainders,  and  at  length  those  of  the  proposed  lines ;  whence 
we  obtain  their  ratio  in  numbers. 

For  example,  if  we  find  GB  is  contained  exactly  twice  in 
FD,  GB  will  be  the  common  measure  of  the  two  proposed 
lines.  Let  GB  be  called  unity,  then  FD  will  be  equal  to  2. 
But  EB  contains  FD  once, plus  GB;  therefore,  EB=3.  CD 
contains  EB  once,  plus  FD  ;  therefore,  CD =5.  AB  contains 
CD  twice,  plus  EB  ;  therefore,  AB  =  13.  Consequently,  the 
ratio  of  the  two  lines  AB,  CD  is  that  of  13  to  5. 

However  far  the  operation  is  continued,  it  is  possible  that 
we  may  never  find  a  remainder  which  is  contained  an  exact 
number  of  times  in  the  preceding  one.  In  such  cases,  the  ex- 


3tf  GEOMETRY. 

act  ratio  can  not  be  expressed  in  numbers  ;  but,  by  taking  the 
measuring  unit  sufficiently  small,  a  ratio  may  always  be 
found,  which  shall  approach  as  near  as  we  please  to  the  true 
ratio. 

So,  also,  in  comparing  two  sur-  unit 
faces,  we  seek  some  unit  of  meas-  Q 
ure  which  is  contained  an  exact 
number  of  times  in  each  of  them. 
Let  A  and  B  represent  two  sur- 
faces, and  let  a  square  inch  be 
the  unit  of  measure.  Now,  if 
this  measuring  unit  is  contained 
15  times  in  A  and  24  times  in  B,  then  the  ratio  of  A  to  B  is 
that  of  15  to  24.  And  although  it  may  be  difficult  to  find 
this  measuring  unit,  \ve  may  still  conceive  it  to  exist ;  or,  if 
there  is  no  unit  which  is  contained  an  exact  number  of  times 
in  both  surfaces,  yet,  since  the  unit  may  be  made  as  small  as 
we  please,  we  may  represent  their  ratio  in  numbers  to  any 
degree  of  accuracy  required. 

Again,  if  we  wish  to  find  the  ratio  of  two  solids,  A  and  B, 
we  seek  some  unit  of  measure  which  is  contained  an  exact 
number  of  times  in  each  of  them.  If  we  take  a  cubic  inch 
as  the  unit  of  measure,  and  we  find  it  to  be  contained  9  times 
in  A,  and  13  times  in  B,  then  the  ratio  of  A  to  B  is  the  same 
as  that  of  9  to  13.  And  even  if  there  is  no  unit  which  is 
contained  an  exact  number  of  times  in  both  solids,  still,  by 
taking  the  unit  sufficiently  small,  we  may  represent  their  ra- 
tio in  numbers  to  any  required  degree  of  precision. 

Hence  the  ratio  of  two  magnitudes  in  geometry,  is  the 
same  as  the  ratio  of  two  numbers,  and  thus  each  magnitude 
has  its  numerical  representative.  We  therefore  conclude 
that  ratio  in  geometry  is  essentially  the  same  as  in  arith- 
metic, and  we  might  refer  to  our  treatise  on  algebra  for  such 
properties  of  ratios  as  we  have  occasion  to  employ.  How- 
ever, in  order  to  render  the  present  treatise  complete  in  it- 
self, we  will  here  demonstrate  the  most  useful  properties. 


Definitions. 

Def.  1.  Ratio  is  the  relation  which  one  magnitude  bears  to 
another  with  respect  to  quantity. 

Thus,  the  ratio  of  a  line  two  inches  in  length,  to  another 
six  inches  in  length  is  denoted  by  2  divided  by  6,  i.  e.,  f  or 
i,  the  number  2  being  the  third  part  of  G.  So,  also,  the  ra- 
tio of  3  feef  to  6  feet  is  expressed  by  £  or  |. 

A  ratio  is  most  conveniently  written  as  a  fraction ;  thus. 


BOOK    II  3 

the  ratio  of  A  to  B  is  written  ^-.     The  two  magnitudes  com 

pared  together  are  called  the  terms  of  the  ratio ;  the  first  is 
called  the  antecedent,  and  the  second  the  consequent. 

Def.  2.  Proportion  is  an  equality  of  ratios. 

Thus,  if  A  has  to  B  the  same  ratio  that  C  has  to  D,  these 
fimr  quantities  form  a  proportion,  and  we  write  it 

A     C 
B=D' 
01  A  :  B  :    C  :  D. 

Tne  hist  and  last  terms  of  a  proportion  are  called  the  two 
extremes,  and  the  second  and  third  terms  the  two  means. 

Of  four  proportional  quantities,  the  last  is  called  a  fourth 
proportional  to  the  other  three,  taken  in  order. 
Since  A_g, 

it  is  obvious  that  if  A  is  greater  than  B,  C  must  be  greater 
than  D  ;  if  equal,  equal ;  and  if  less,  less  ;  that  is,  if  one  ante- 
cedent is  greater  than  its  consequent,  the  other  antecedent 
must  be  greater  than  its  consequent ;  if  equal,  equal ;  and  if 
less,  less. 

Def.  3.  Three  quantities  are  said  to  be  proportional,  when 
the  ratio  of  the  first  to  the  second  is  equal  to  the  ratio  of  the 
second  to  the  third ;  thus,  if  A,  B,  and  C  are  in  proportion, 
then 

A  :  B  :  :  B  :  C. 

In  this  case  the  middle  term  is  said  to  be  a  mean  propo? 
tional  between  the  other  two. 

Def.  4.  Two  magnitudes  are  said  to  be  equimultiples  ol 
two  others,  when  they  contain  those  others  the  same  number 
of  times  exactly.  Thus,  7 A,  7B  are  equimultiples  of  A  and 
B  ;  so,  also,  are  mA.  and  mE. 

Def.  5.  The  ratio  of  B  to  A  is  said  to  be  the  reciprocal  of 
the  ratio  of  A  to  B. 

Def.  6.  Inversion  is  when  the  antecedent  is  made  the  con- 
sequent, and  the  consequent  the  antecedent. 

Thus,  if  A  :  B  :  •  C  :  D ; 

then,  inversely, 

B  :  A  .  :  D  :  C. 

Def.  7.  Alternation  is  when  antecedent  is  compared  with 
antecedent,  and  consequent  with  consequent 

Thus,  if  A  :  B  :  :  C  :  D ; 

then,  by  alternation, 

A  :  C  :  :  B  :  D. 

Def.  8.  Composition  is  when  the  sum  of  antecedent  ana 
consequent  is  compared  either  with  the  antecedent  or  con 
seouent. 


38  GEOMETRY 

Thus,  if  A  :  B  :  :  C  :  D ; 

then,  by  composition, 

A+B  :  A  :  :  C-f  D  :  C,  and  A+B  :  B  :  :  C+D  :  D. 
Def.  9.  Division  is  when  the  difference  of  antecedent  ana 
consequent  is  compared  either  with  the  antecedent  or  con 
sequent. 

Thus,  if  A  :  B  :  :  C  :  D ; 

fhen,  by  division, 

A— B  :  A  :  :  C— D  :  C,  and  A— B  :  B  :  :  C— D  :  D. 


Axioms. 

1.  Equimultiples  of  the  same,  or  equal  magnitudes,  are 
equal  to  each  other. 

2.  Those  magnitudes  of  which  the  same  or  equal  magni- 
tudes are  equimultiples,  are  equal  to  each  other. 


PROPOSITION    I.       THEOREM. 

If  four  quantities  are  proportional,  the  product  of  the  two  ex- 
°,mes  is  equal  to  the  product  of  the  two  means. 

It  has  been  shown  that  the  ratio  of  two  magnitudes,  wheth- 
er they  are  lines,  surfaces,  or  solids,  is  the  same  as  that  of 
.  wo  numbers,  which  we  call  their  numerical  representatives. 

Let,  then,  A,  B,  C,  D  be  the  numerical  representatives  of 
four  proportional  quantities,  so  that  A  :  B  :  :  C  :  D  ;  the'.n 
will  AxD=BxC. 

For,  since  the  four  quantities  are  proportional, 

A__C 

B~D* 

Multiplying  each  of  these  equal  quantities  by  B  (Axiom  1) 
we  obtain 


Multiplying  each  of  these  last  equals  by  D,  we  have 

AxD=BxC. 

Cor.  If  there  are  three  proportional  quantities,  the  product 
of  the  two  extremes  is  equal  to  the  square  of  the  mean. 

Thus,  if  A  :  B  :  :  B  :  C  ; 

*hen,  by  ,he  proposition, 

,  which  is  equa  to  Ba. 


BOOK    II.  3t 


PROPOSITION  ii.     THEOREM  (Converse  of  Prop.  /.). 

If  the  product  of  two  quantities  is  equal  to  the  product  of  twc 
other  quantities,  the  first  two  may  be  made  the  extremes,  and 
the  other  two  the  means  of  a  proportion. 

Thus,  suppose  we  have  AxD=BxC  ;  then  will 

A  :  B  :  :  C  :  D. 

For,  since  AxD=BxC,  dividing  each  of  these  equals  by 
D  (Axiom  2),  we  have 


Dividing  each  of  these  last  equals  by  B,  we  obtain 
AC 


that  is,  the  ratio  of  A  to  B  is  equal  to  that  of  C  to  D, 
or,  A  :  B  :  :  C  :  D. 


PROPOSITION    III.       THEOREM. 

If  four  quantities  are  proportional,  they  are  also  proportion- 
al when  taken  alternately. 

Let  A,  B,  C,  D  be  the  numerical  representatives  of  foui 
proportional  quantities,  so  that  A  :  B  :  :  C  :  D ;  then  will 
A  :  C  :  :  B  :  D. 

For,  since  A  :  B  :  :  C  :  D, 

by  Prop.  L,  AxD=BxC. 

And,  since  A  X  D =B  X  C, 

bv  Prop.  II.,  A  :  C  :  :  B  :  D. 


PROPOSITION   IV.       THEOREM. 

Ratios  that  are  equal  to  the  same  ratio,  are  equal  to  each 
other. 

Let  A  :  B  :  :  C  :  D, 

and  A  :  B  :  :  E  :  F ; 

then  will  C  :  D  :  :  E  :  F. 

For,  since  A  :  B  :  :  C  :  D, 

.  A     C 

we  have  T7=?V 

r>      \J 


40  GEOMETRY. 

And,  since  A  :  B  :  :  E    F, 

A    E 
we  have  B=F* 

C         E  A 

But  =:  and  ™  being  severally  equal  to  ^  must  be  equal  to 

each  other,  and  therefore 

C  :  D  :  :  E  :  F. 

Cor.  If  the  antecedents  of  one  proportion  are  equai  to  the 
antecedents  of  another  proportion,  the  consequents  are  pro 
portional. 

If  A  :  B  •  :  C  :  D, 

and  A:E::C:F; 

then  will  B  :  D  :  :  E  :  P. 

For,  by  alternation  (Prop.  III.),  the  first  proportion  b&« 
comes 

A  :  C  :  :  B  :  D, 

and  the  second,  A  :  C  :  :  E  :  F. 

Therefore,  by.  the  proposition, 

EL;  D  :  :  E  :  F. 


PROPOSITION  V.       THEOREM. 

If  four  quantities  are  proportional,  they  are  also  proportion 
al  when  taken  inversely. 


Let  A 

then  will  B 

For,  since  A 


:  C  :  D ; 

:D:C. 

:C:D, 


bvProp.  I.,  AxD=BxC, 

or,  BXC=AXD; 

therefore,  by  Prop.  II., 

B  :  A  :  :  D  :  C. 


PROPOSITION   VI.       THEOREM. 

If  four  quantities  are  proportional,  they  are  also  proportion 
al  by  composition. 

Let  A  :  B  :  :  C  :  D, 

hen  will  A-f  B  :  A  :  :  C-f  D  .  C. 

For,  since  A  :  B  :  :  C  :  D, 

by  Prop.  I.,  BxC=AxD. 

To  each  of  these  equals  add 

AxC=AxC, 
then  AxC-fBxC=AxC+AxD, 


BOOK    II. 


(A+B)xC=Ax(C4-D; 

Therefore,  by  Prop.  II., 

A+B  :  A  :  :  C+D  :  C. 


PROPOSITION   VII.       THEOREM. 

If  four  quantities  are  proportional,  they  are  also  proportion 
At  by  division. 

Let  A  :  B      C  :  D ; 

then  wiL  A— B  :  A  :  :  C— D  :  C. 

For,  since  A  :  B  :  :  C  :  D, 

by  Prop.  I.,  BxC=AxD. 

Subtract  each  of  these  equals  from  AxC; 
then  AxC— BxC=AxC— AxD, 

or,  (A— B)xC=Ax(C  — D). 

Therefore,  by  Prop.  II., 

A— B:  A::C  — D  :  C. 

Cor.  A+B  :  A— B  :  :  C+D  :  C— D. 


PROPOSITION    VIII.       THEOREM. 

Equimultiples  of  two  quantities  have  the  same  ratio  as  trie 
quantities  themselves. 

Let  A  and  B  be  any  two  quantities,  and  mA,  mE  then 
equimultiples ;  then  will 

A  f  B  :  :  mA  :  mE. 

For  7/zx  AxB=7?zx  AxB, 

or,  AxmE  — E*mA. 

Therefore,  by  Prop.  II., 

A  :  B  :  :  mA  :  mE. 


PROPOSITION    IX.       THEOREM. 

If  any  number  of  quantities  are  proportional,  any  one  ante 
&dent  is  to  its  consequent,  as  the  sum  of  all  the  antecedents.  11 
to  the  sum  of  all  the  consequents. 

Let  A  :  B  :  :  C  :  D  :  :  E  :  F,  &c. ; 

then  will  A  :  B  :  :  A+C+E  :  B+D+F 

For,  since  A  :  B  :  :  C  :  D, 

we  have  AxD=BxC. 

And,  since  A  :  B  :  :  E  :  F, 

we  have  AxF=BxE. 

To  these  equals  a^d 

AxB=AxB. 


42  GEOMETRY 

and  we  have 

AxB+AxD+AxF=AxB+BxC+BxE 
or,  Ax(B+D+F)=Bx(A+ClE). 

Therefore,  by  Prop.  II., 

A  :  B  : :  A+C+E  :  B+D+F. 


PROPOSITION  X.       THEOREM. 

If  four  quantities  are  proportional,  their  squares  or  cubes 
are  also  proportional. 

Let  A  :  B  :  :  C  :  D ; 

then  will  A2  :  B2  :  :  Ca :  D2, 

and  A8  :  B3  :  :  C8  :  D3. 

For,  since  A  :  B  :  :  C  :  D, 

ny  Pro'p.  L,  AxD=BxC; 

or,  multiplying  each  of  these  equals  by  itself  (Axiom  1),  we 
have 

A2xD2=B2xC2; 
and  multiplying  these  last  equals  by  AxD  =  BxC,  we  have 

A3xD3=B8xCs. 
Therefoie,  by  Prop.  II., 

A2 :  B2 :  :  C2 :  D9, 
and  A9  :  B8  :  :  C3  :  D3. 


PROPOSITION    XI.       THEOREM. 

If  there  are  two  sets  of  proportional  quantities,  the  product* 
oj  the  corresponding  terms  are  proportional. 

Let  A  :  B  :  :  C  :  D, 

and  E  :  F  :  :  G  :  H ; 

then  will  AxE  :  BxF  :  :  CxG  :  DxH. 

For,  since  A  :  B  :  :  C  :  D, 

by  Prop.  L,  AxD=BxC. 

And,  since  E  :  F  :  :  G  :  H, 

LyProp.  L,  ExH^FxG. 

Multiplying  together  these  equal  quantities,  we  have 

AxDxExH=BxCxFxG; 
or,  (AxE)x(DxH)  =  (BxF)x(CxG); 

therefore,  by  Prop.  II., 

AxE:BxF::CxG:DxH. 

Cor.  If  A  :  B  :  :  C  :  D, 

and  B  :  F  :  :  G  :  H  ; 

then  A  :F::  CxG:  DxH. 


BOOK    II.  43 


Foi,  by  the  proposition, 

AxB:BxF::CxG:DxH  . 
Also,  by  Prop.  VIIL, 

AxB :BxF:  :  A : F; 
hence,  by  Prop.  IV., 

A  :F::CxG:DxH. 


PROPOSITION    XII.       THEOREM. 

If  three  quantities  are  proportional,  the  first  is  to  the  /Aird, 
as  the  square  of  the  first  to  the  square  of  the  second. 

Thus,  if  A:B::B    :C; 

then  A  :  C  :  :  Aa :  Ba. 

For,  since  A  :  B  :  :  B   :  C, 

and  A  :  B  :  :  A  :  B ; 

therefore,  by  Prop.  XL, 

A3:B2::  AxB:BxC. 
But,  by  F;op.  VIIL, 

AxB:BxC:: A:C; 
hence,  by  Prop.  IV  ,    A  :  C  : :  Aa :  Bs, 


44  GEOMETRY 


BOOK  III 

THE  CIRCLE,  AND  THE  MEASURE  OF  ANGLES. 
Definitions. 

1.  A  circle  is  a  plane  figure  bounded  by  a  line,  every  point 
of  which  is  equally  distant  from  a  point  within,  called  the 
center. 

This  bounding  line  is  called  the  circumfer- 
ence of  the  circle. 

2.  A  radius  of  a  circle  is  a  straight  line 
drawn  from  the  center  to  the  circumference. 

A  diameter  of  a  circle  is  a  straight  line 
passing  through  the  center,  and  terminated 
both  ways  by  the  circumference. 

Cor.  All  the  radii*  of  a  circle  are  equal ;  all  the  diameters 
are  equal  also,  and  each  double  of  the  radius. 

3.  An  arc  of  a  circle  is  any  part  of  the  circumference. 
The  chord  of  an  arc  is  the  straight  line  which  joins  its  two 

extremities. 

4.  A  segment  of  a  circle  is  the  figure  included  between  an 
arc  and  its  chord. 

5.  A  sector  of  a  circle  is  the  figure  included  between  an 
arc,  and  the  two  radii  draw,n  to  the  extremities  of  the 'arc. 

6.  A  straight  line  is  said  to  be  inscribed  in  a  circle,  when 
its  extremities  are  on  the  circumference. 

An  inscribed  angle  is  one  whose  sides  are 
inscribed. 

7.  A  polygon  is  said  to  be  inscribed  in  a 
c  rcle  when  all  its  sides  are  inscribed.     The 
circle  is  then  said  to  be  described  about  the 
polygon. 

8.  A  secant  is  a  line  which  cuts  the  cir- 
cumference, and  lies  partly  within  and  partly  without  the 
circle. 

9.  A  straight  line  is  said  to  touch  a  "circle,  when  it  meets 
the  circumference,  and,  being   produced,  does   not   cut  it. 
Such  a  line  is  called  a  tangent,  and  the  point  in  which  i* 
meets  the  circumference,  is  called  the  point  of  contact. 


BOOK    III. 


45 


10.  Two  circumferences  touch  each 
other  when  they  meet,  but  do  not  cut 
one  another. 


11.  A  polygon  is  described  about  a  circle, 
when  each  side  of  the  polygon  touches  the  cir- 
cumference of  the  circle. 

In  the  same  case,  the  circle  is  said  to  be  in- 
scribed in  the  polygon. 


PROPOSITION  I.       THEOREM. 

Every  diameter  divides  the  circle  and  its  circumference  into 
two  equal  parts. 

Let  ACBD  be  a  circle,  and  AB  its  di- 
ameter. The  line  AB  divides  the  circle 
and  its  circumference  into  two  equal  parts. 
For,  if  the  figure  ADB  be  applied  to  the 
figure  ACB,  while  the  line  AB  remains 
common  to  both,  the  curve  line  ACB  must 
coincide  exactly  with  the  curve  line  ADB. 
For,  if  any  part  of  the  curve  ACB  were  to 
fall  either  within  or  without  the  curve  ADB,  there  would  be 
points  in  one  or  the  other  unequally  distant  from  the  center 
which  is  contrary  to  the  definition  of  a  circle.  Therefore 
every  diameter,  &c. 


PROPOSITION  II.       THEOREM. 

A  straight  line  can  not  meet  the  circumference  of  a  circle  t* 
more  than  two  points. 

For,  if  it  is  possible,  let  the  straight 
line  ADB  meet  the  circumference  CDE 
in  three  points,  C,  D,  E.  Take  F,  the 
center  of  the  circle,  and  join  FC,  FD, 
FE.  Then,  because  F  is  the  center  of 
the  circle,  the  three  straight  lines  FC, 
FD,  FE  are  all  equal  to  each  other; 
hence,  three  equal  straight  lines  have 
been  drawn  from  the  same  point  to  the  same  straight  line. 


46  GEOMETRY. 

which  is  impossible  (Prop.  XVII.,  Cor.  2,  Book  I.\     There- 
fore, a  straight  line,  &c. 


PROPOSITION   III.       THEOREM. 

in  equal  circles,  equal  arcs  are  subtended  by  equal  chords 
and,  conversely,  equal  chords  subtend  equal  arcs. 

Let    ADB,  EHF    be 

equal  circles,  and  let  the 
arcs  AID,  EMH  also  be 
equal ;    then     will     the  A[ 
chord  AD   be  equal   to 
the  chord  EH. 

For,  the  diameter  AB 
being  equal  to  the  diameter  EF,  the  semicircle  ADB  may  be 
applied  exactly  to  the  semicircle  EHF,  and  the  curve  line 
AIDB  will  coincide  entirely  with  the  curve  fine  EMHF 
(Prop.  L).  But  the  arc  AID  is,  by  hypothesis,  equal  to  the 
arc  EMH ;  hence  the  point  D  will  fall  on  the  point  H,  and 
therefore  the  chord  AD  is  equal  to  the  chord  EH  (Axiom 
11,  B.  L). 

Conversely,  if  the  chord  AD  is  equal  to  the  chord  EH,  then 
the  arc  AID  will  be  equal  to  the  arc  EMH. 

For,  if  the  radii  CD,  GH  are  drawn,  the  two  triangles 
ACD,  EGH  will  have  their  three  sides  equal,  each  to  each 
viz. :  AC  to  EG,  CD  to  GH,  and  AD  equal  to  EH  ;  the  tri 
angles  are  consequently  equal  (Prop.  XV.,  B.  I.),  and  the  an 
gle  ACD  is  equal  to  the  angle  EGH.  Let,  now,  the  semicir- 
cle ADB  be  applied  to  the  semicircle  EHF,  so  that  AC  may 
coincide  with  EG ;  then,  since  the  angle  ACD  is  equal  to  the 
angle  EGH,  the  radius  CD  will  coincide  with  the  radius  GH, 
and  the  point  D  with  the  point  H.  Therefore,  the  arc  AID 
must  coincide  with  the  arc  EMH,  and  be  equal  to  it.  Hence, 
in  equal  circles,  &c. 


PROPOSITION  IV.       THEOREM. 

In  equal  circles,  equal  angles  at  the  center,  are  subtended  by 
equal  arcs  ;  and,  conversely,  equal  arcs  subtend  equal  angles  at 
the  center. 

Let  AGB,  DHE  be  two  equal  circles,  and  let  ACB,  DFE 
be  equal  angles  at  their  centers ;  then  will  the  arc  AB  be 
equal  to  the  arc  DE.  Join  AB,  DE  ;  and,  because  the  cir 


JOOK    III. 


cles  AGB,  DHE  are  equal,  their 

radii  are  equal.     Therefore,  the 

two  sides  CA,  CB  are  equal  to 

the  two  sides  FD,  FE ;  also,  the 

angle  at  C  is  equal  to  the  angle 

at  F ;  therefore,  the  base  AB  is 

equal  to  the  base  DE  (Prop.  VI., 

B.  I.).     And,  because  the  chord  AB 

is  equal  to  the  chord  DE,  the  arc  AB  must  be  equal  to  the 

arc  DE  (Prop.  III.). 

Conversely,  if  the  arc  AB  is  equal  to  the  arc  DE,  the  an- 
gle ACB  will  be  equal  to  the  angle  DFE.  For,  if  these  an- 
gles are  not  equal,  one  of  them  is  the  greater.  Let  ACB  be 
the  greater,  and  take  ACI  equal  to  DFE ;  then,  because 
equal  angles  at  the  center  are  subtended  by  equal  arcs,  the 
arc  AI  is  equal  to  the  arc  DE.  But  the  arc  AB  is  equal  to 
the  arc  DE ;  therefore,  the  arc  AI  is  equal  to  the  arc  AB, 
the  less  to  the  greater,  which  is  impossible.  Hence  the  an- 
gle ACB  is  not  unequal  to  the  angle  DFE,  that  is,  it  is  equa* 
to  it.  Therefore,  in  equal  circles,  &c. 


PROPOSITION   V.       THEOREM. 

In  the  same  circle,  or  in  equal  circles,  a  greater  arc  is  sub 
tended  by  a  greater  chord;  and,  conversely,  the  greater  chord 
subtends  the  greater  arc. 

In  the  circle  AEB,  let  the  arc  AE  be 
greater  than  the  arc  AD ;  then  will  the 
chord  AE  be  greater  than  the  chord  AD. 

Draw  the  radii  CA,  CD,  CE.  Now,  if 
the  arc  AE  were  equal  to  the  arc  AD, 
the  angle  ACE  would  be  equal  to  the  an- 
gle ACD  (Prop.  IV.)  ;  hence  it  is  clear 
that  if  the  arc  AE  be  greater  than  the  arc 
AD,  the  angle  ACE  must  be  greater  than  the  angle  ACD. 
But  the  two  sides  AC,  CE  of  the  triangle  ACE  are  equal  to 
the  two  AC,  CD  of  the  triangle  ACD,  and  the  angle  ACE  is 
greater  than  the  angle  ACD ;  therefore,  the  third  side  AE  is 
greater  than  the  third  side  AD  (Prop.  XIII. ,  B.  I.)  ;  hence 
the  chord  which  subtends  the  greater  arc  is  the  greater. 

Conversely,  if  the  chord  AE  is  greater  than  the  chord  AD 
the  arc  AE  is  greater  than  the  arc  AD.  For,  because  the 
two  triangles  ACE,  ACD  have  two  sides  of  the  one  equal 
to  two  sides  of  the  other,  each  to  each,  but  the  base  AE  of 
the  one  is  greater  than  the  base  AD  of  the  other,  therefore 


48  GEOMETRF 

the  angle  ACE  is  greater  than  the  angle  ACD  (Prop.  Xl  V . 
B.  I.)  ;  and  hence  the  arc  .AE  is  greater  than  the  arc  AD 
(Pi-op.  IV.).  Therefore,  in  the  same  circle,  &c. 

Scholium,  The  arcs  here  treated  of  are  supposed  to  be 
less  than  a  semicircumference.  If  they  were  greater,  the  op- 
posite property  would  hold  true,  that  is,  the  greater  the  arc 
the  smaller  the  chord. 


PROPOSITION    VI.       THEOREM. 

The  radius  which  is  perpendicular  to  a  chord,  bisects 
chord,  and  also  the  arc  which  it  subtends. 

Let  ABG  be  a  circle,  of  which  AB  is  a 
chord,  and  CE  a  radius  perpendicular  to 
it;  the  chord  AB  will  be  bisected  in  D, 
and  the  arc  AEB  will  be  bisected  in  E. 

Draw  the  radii  C  A,  CB.  The  two  right- 
angled  triangles  CDA,  CDB  have  the  side 
AC  equal  to  CB,  and  CD  common ;  there- 
fore the  triangles  are  equal,  and  the  base 
AD  is  equal  to  the  base  DB  (Prop.  XIX., 
B.  I.). 

Secondly,  since  ACB  is  an  isosceles  triangle,  and  the  line 
CD  bisects  the  base  at  right  angles,  it  bisects  also  the  verti- 
cal angle  ACB  (Prop.  X.,  Cor.  1,  B.  I.).  And,  since  the  an- 
gle ACE  is  equal  to  the  angle  BCE,  the  arc  AE  must  be 
equal  to  the  arc  BE  (Prop.  IV.)  ;  hence  the  radius  CE,  per- 
pendicular to  the  chord  AB,  divides  the  arc  subtended  by 
this  chord,  into  two  equal  parts  in  the  point  E.  Therefore, 
the  radius,  &c. 

Scholium.  The  center  C,  the  middle  point  D  of  the  chord 
AB,  and  the  middle  point  E  of  the  arc  subtended  by  this 
chord,  are  three  points  situated  in  a  straight  line  perpendic- 
ular to  the  chord.  Now  two  points  are  sufficient  to  deter- 
mine the  position  of  a  straight  line ;  therefore  any  straight 
jae  which  passes  through  two  of  these  points,  will  necessari- 
iy  pass  through  the  third,  and  be  perpendicular  to  the  chord. 
Also,  the  perpendicular  at  the  middle  of  a  chord  passes  through 
the  center  of  the  circle,  and  through  {he  middle  of  tlw  arc 
ten  did  by  the  chord. 


BOOK    111.  49 


PROPOSITION    VII.       THEOREM. 

Through  three  given  points,  not  in  the  same  straight  lint 
cue  circumference  may*be  made  to  pass,  and  but  one. 

Let  A,  B,  C  be  three  points  not  in  the  same  straight  line , 
they  all  lie  in  the  circumference  of  the  same  circle.  Join 
AB,  AC,  arid  bisect  these  lines  by  the 
perpendiculars  DF,  EF;  DF  and  EF 
produced  wLI  meet  one  another.  For, 
join  DE  ;  then,  because  the  angles  ADF, 
AEF  are  together  equal  to  two  right  an- 
gles,  the  angles  FDE  and  FED  are  to- 
gether  less  than  two  right  angles  ;  there- 
fore DF  and  EF  will  meet  if  produced 
(Prop.  XXIIL,  Cor.  2,  B.  I.).  Let  them 
meet  in  F.  Since  this  point  lies  in  the  perpendicular  DF,  it  is 
equally  distant  from  the  two  points  A  and  B  (Prop.  XVIII.,' 
B.  I.) ;  and,  since  it  lies  in  the  perpendicular  EF,  it  is  equally 
distant  from  the  two  points  A  and  C ;  therefore  the  throe 
distances  FA,  FB,  FC  are  all  equal;  hence  the  circumfe- 
rence described  from  the  center  F  with  the  radius  FA  will 
pass  through  the  three  given  points  A,  B,  C. 

Secondly.  No  other  circumference  can  pass  through  the 
same  points.  For,  if  there  were  a  second,  its  center  could 
not  be  out  of  the  line  DF,  for  then  it  would  be  unequally  dis- 
tant from  A  and  B  (Prop.  XVIIL,  B.  I.) ;  neither  could  it  be 
out  of  the  line  FE,  for  the  same  reason  ;  therefore,  it  must  be 
on  both  the  lines  DF,  FE.  But  two  straight  lines  can  not 
cut  each  other  in  more  than  one  point ;  hence  only  one  cir- 
cumference can  pass  through  three  given  points.  Therefore, 
through  three  given  points,  &c. 

Cor.  Two  circumferences  can  not  cut  each  other  in  more 
than  two  points,  for,  if  they  had  three  common  points,  they 
would  have  the  same  center,  and  would  coincide  with  each 
other. 


PROPOSITION   VIII.       THEOREM. 

Equal  chords  are  equally  distant  from  the  center  ;  and  of  two 
unequal  chords,  the  less  is  the  more  remote  from  the  center. 

Let  the  chords  AB,  DE,  in  the  circle  ABED,  be  equal  to 
;>ne  another  ;  they  are  equally  distant  from  the  centei      T°.ke 

C 


50 


GEOMETRV. 


C,  the  center  of  the  circle,  and  from  it 

draw  CF,  CG,  perpendiculars  to  AB, 

DE.     Join  CA,  3D  ;  then,  because  the 

radius  CF  is  perpendicular  to  the  chord 

AB,  it  bisects  it  (Prop.  VI.).      Hence 

AF  is  the  half  of  AB  ;  and,  for  the  same 

reason,  DG  is  the  half  of  DE.     But  AB 

is  equal  to  DE ;  therefore  AF  is  equal 

to  DG  (Axiom  7,  B.  I.).     Now,  in. the 

right-angled  triangles  ACF,  DCG,  the  hypothenuse  AC  is 

equal  to  the  hypothenuse  DC,  and  the  side  AF  is  equal  to 

tha  side  DG;  therefore  the  triangles  are  equal,  and  CF  is 

equal  to  CG  (Prop.  XIX.,  B.  I.) ;  hence  the  two  equal  chords 

AB,  DE  are  equally  distant  from  the  center. 

Secondly.  Let  the  chord  AH  be  greater  than  the  chord  DE  ; 
DE  is  further  from  the  center  than  AH.  For,  because  the 
chord  AH  is  greater  than  the  chord  DE,  the  arc  ABH  is 
greater  than  the  arc  DE  (Prop.  V.).  From  the  arc  ABH 
cut  off  a  part,  AB,  equal  to  DE  ;  draw  the  chord  AB,  and 
•let  fall  CF  perpendicular  to  this  chord,  and  CI  perpendicular 
to  AH.  It  is  plain  that  CF  is  greater  than  CK,  and  CK 
than  CI  (Prop.  XVII.,  B.  I.) ;  much  more,  then,  is  CF  great- 
_er  than  CI.  But  CF  is  equal  to  CG,  because  the  chords  AB, 
DE  are  equal ;  hence  CG  is  greater  than  CI.  Therefore, 
equal  chords,  &c. 

Cor.  Hence  the  diameter  is  the  longest  line  that  can  be  in 
scribed  in  a  circle. 


PROPOSITION   IX.       THEOREM. 

A  straight  line  perpendicular  to  a  diameter  at  its  extremity, 
ts  a  tangent  to  the  circumference. 

Let  ABG  be  a  circle,  the  center  of  which  is  C,  and  the  di- 
ameter AB ;  and  let  AD  be  drawn  from  A  perpendicular  to 
AB  ;  AD  will  be  a  tangent  to  the  circum- 
ference. 

In  AD  take  any  point  E,  and  join 
CE ;  then,  since  CE  is  an  oblique  line, 
it  is  longer  than  the  perpendicular  CA 
(Prop.  XVIL,  B.  I.).  Now  CA  is  equal 
to  CK  ;  therefore  CE  is  greater  than 
CK,  and  the  point  E  must  be  without 
Jie  circle.  But  E  is  any  point  whatev- 
er in  the  line  AD ;  therefore  AD  has 
only  the  point  A  in  common  with  the 


flOOR.    III. 


51 


cncumference,  hence  it  is   a  tangent  (Def.  9).     Therefore, 
a  straight  line,  &c. 

Scholium.  Through  the  same  point  A  in  the  ciicumfer- 
ence,  only  one  tangent  can  be  drawn.  For,  if  possible  let  a 
second  tangent,  AF,  be  drawn ;  then,  since  CA  can  not  be 
perpendicular  to  AF  (Prop.  XVI.,  Cor.,  B.  I.),  another  line, 
CH,  must  be  perpendicular  to  AF,  and  therefore  CH  must  be 
less  than  CA  (Prop.  XVII.,  B.  I. ;  hence  the  point  H  falls 
within  the  circle,  and  AH  produced  will  cut  the  circumfer- 
ence. 


PROPOSITION    X.       THEOREM. 


Two  parallels  intercept  equal  arcs  on  the  circumference. 

The  proposition  admits  of  three  cases: 

First.  When  the  two  parallels  are  se- 
cants, as  AB,  DE.  Draw  the  radius  CH 
perpendicular  to  AB  ;  it  will  also  be  per- 
pendicular to  DE  (Prop.  XXIII.,  Cor. 
1,  B.  I.)  ;  therefore,  the  point  H  will  be 
at  the  same  time  the  middle  of  the  arc 
AHB,  and  of  the  arc  DHE  (Prop.  VI.). 
Hence  the  arc  DH  is  equal  to  the  arc 
HE,  and  the  arc  AH  equal  to  HB,  and  therefore  the  arc  AD 
is  equal  to  the  arc  BE  (Axiom  3,  B.  I.). 

Second.  When  one  of  the  two  par-  _. 

allels  is  a  secant,  and  the  other  a  tan-  D ^-..  *^ E 

gent.  To  the  point  of  contact,  H,  _ 
draw  the  radius  CH ;  it  will  be  per- 
pendicular to  the  tangent  DE  (Prop. 
IX.),  and  also  to  its  parallel  AB.  But 
since  CH  is  perpendicular  to  the  chord 
AB,  the  point  H  is  the  middle  of  the 
arc  AHB  (Prop.  VI.)  ;  therefore  the 
arcs  AH,  HB,  included  between  the 
parallels  AB,  DE,  are  equal. 

Third.  If  the  two  parallels  DE,  FG  are  tangents,  the  one 
at  H,  the  other  at  K,  draw  the  parallel  secant  AB ;  then,  ac- 
cording to  the  former  case,  the  arc  AH  is  equal  to  HB,  and 
the  arc  AK  is  equal  to  KB ;  hence  the  whole  arc  HAK  is 
equal  to  the  whole  arc  HBK  (Axiom  2,  B.  L).  It  is  also  ev- 
ident that  each  of  these  arcs  is  a  semicircumrerence.  There 
fore,  two  parallels,  &c. 


GEOMETRY. 


PROPOSITION    XI.       THEOREM. 

If  two  circumferences  cut  each  other,  the  chord  which  joins 
the  points  of  intersection,  is  bisected  at  right  angles  by  th& 
straight  line  joining  their  centers. 

Let  two  circum- 
ferences cut  each 
other  in  the  points  A 
and  B  ;  then  will  the 
ine  AB  be  a  com- 
Tion  chord  to  the 
two  circles.  Now,  if 
a  perpendicular  be 
erected  from  the  middle  of  this  chord,  it  will  pass  through  C 
and  D,  the  centers  of  the  two  circles  (Prop.  VI.,  Schol.). 
But  only  one  straight  line  can  be  drawn  through  two  given 
points ;  therefore,  the  straight  line  which  passes  through  thft 
centers,  will  bisect  the  common  chord  at  right  angles. 


PROPOSITION    XII.       THEOREM. 


If  two  circumferences  touch  each  other,  either  externally  01 
internally,  the  distance  of  their  centers  must  be  equal  to  the 
turn  or  difference  of  their  radii. 

It  is  plain  that  the  centers  of  the  circles  and  the  point  of 


contact  are  in  the  same  straight  line ;  for,  if  possible,  let  the 
point  of  contact,  A,  be  without  the  straight  line  CD.  From 
A  let  fall  upon  CD,  or  CD  produced,  the  perpendicular  AE, 
and  produce  it  to  B,  making  BE  equal  to  AE.  Then,  in  the 
triangles  ACE,  BCE,  the  side  AE  is  equal  to  EB,  CE  is  com- 
mon, and  the  angle  AEC  is  equal  to  the  angle  BEC ;  there- 
fore AC  is  equal  to  CB  (Prop.  VI.,  B.  I.),  and  the  point  B  is 
in  the  circumference  ABF.  In  the  same  manner,  it  may  be 
*hown  to  be  in  the  circumference  ABG,  and  hence  the  point 


BOOK  m. 


53 


15  is  in  both  circumferences.  Therefore  the  two  circumfe- 
rences have  two  points,  A  and  B,  in  common ;  that  is,  they  cut 
each  other,  which  is  contrary  to  the  hypothesis.  Therefore, 
the  point  of  contact  can  not  be  without  the  line  ^pining  the 
centers  ;  and  hence,  when  the  circles  touch  each  other  exter- 
nally, the  distance  of  the  centers  CD  is  equal  to  the  sum  of 
the  radii  CA,  DA;  and  when  they  touch  internally,  the  dis 
tance  CD  is  equal  to  the  difference  of  the  radii  CA,  DA 
Therefore,  if  two  circumferences,  &c. 

SchoL  If  two  circumferences  touch  each  other,  externally 
or  internally,  their  centers  and  the  point  of  contact  are  in 
the  same  straight  line. 


PROPOSITION    XIII.       THEOREM. 

If  two  circumferences  cut  each  other,  the  distance  between 
their  centers  is  less  than  the  sum  of  their  radii,  and  greater 
than  their  difference. 

Let  two  circumferences  cut  each 
other  in  the  point  A.  Draw  the  ra- 
dii C A,  DA  ; .  then,  because  any  two 
sides  of  a  triangle  are  together  great- 
er than  the  third  side  (Prop.  VIII.,  B. 
I.),  CD  must  be  less  than  the  sum  of 
AD  and  AC.  Also,  DA  must  be  less 
than  the  sum  of  CD  and  CA ;  or,  subtracting  CA  from  these 
unequals  (Axiom  5,  B.  I.),  CD  must  be  greater  than  the  dif- 
ference between  DA  and  CA.  Therefore,  if  two  circumfe- 
rences, &c. 


PROPOSITION    XIV.       THEOREM. 

In  equal  circles,  angles  at  the  center  have  the  same  ratit 
with  the  intercepted  arcs. 

Case  first.  When 
tiie  angles  are  in  the 
ratio  of  two  whole 
numbers. 

Let  ABG,  DFH 
be  equal  circles,  and 
let  the  angles  ACB, 
DEF  at  their  cen- 
ters be  in  the  ratio  of  two  whole  numbers  ;  then  will 
the  angle  ACB  :  angle  DEF  :  :  arc  AF  :  arc  DF. 


54  GEOMETRY. 

Suppose,  foi  exan\ple,  that  the  angles  ACB,  DEF  are  to 
each  other  as  7  to  4 ;  or,  which  is  the  same  thing,  suppose 
that  the  angle  M,  which  may  serve  as  a  common  measure, 
is  contained  seven  times  in  the  angle  ACB,  and  four  times  in 
the  angle  DEF.  The  seven  partial  angles  into  which  ACB 
is  divided,  being  aach  equal  to  any  of  the  four  partial  angles 
into  which  DEF  is  divided,  the  partial  arcs  will  also  be 
equal  to  each  other  (Prop.  IV.),  and  he  entire  arc  AB  will 
be  to  the  entire  arc  DF  as  7  to  4.  Now  the  same  reasoning 
would  apply,  if  in  place  of  7  and  4  any  whole  numbers  what- 
ever were  employed  ;  therefore,  if  the  ratio  of  the  angles 
ACB,  DEF  can  be  expressed  in  whole  numbers,  the  arcs  AB, 
DF  will  be  to  each  other  as  the  angles  ACB,  DEF. 

Case  second.  When  the  ratio  of  the  angles  can  not  be  ex 
pressed  by  whole  numbers. 

Let  ACB,  ACD  be  two  an- 
gles having  any  ratio  whatev- 
er. Suppose  ACD  to  be  the 
smaller  angle,  and  let  it  be 
placed  on  the  greater;  then 
will  the  angle  ACB  :  angle 
ACD  :  :  arc  AB  :  arc  AD. 

For,  if  this  proportion  is  not  true,  the  first  three  terms  re- 
maining the  same,  the  fourth  must  be  greater  or  less  than 
AD.  Suppose  it  to  be  greater,  and  that  we  have 

Angle  ACB  :  angle  ACD  :  :  arc  AB  :  arc  AI. 

Conceive  the  arc  AB  to  be  divided  into  equal  parts,  each 
(ess  than  DI ;  there  will  be  at  least  one  point  of  division  be- 
Iweeu  D  and  I.  Let  II  be  that  point,  and  join  CH.  The 
arcs  AB,  AH  will  be  to  each  other  in  the  ratio  of  two  whole 
numbers,  and,  by  the  preceding  case,  we  shall  have 

Angle  ACB  :  angle  ACH  :  :  arc  AB  :  arc  AH. 

Comparing  these  two  proportions  with  each  other,  and  ob- 
serving that  the  antecedents  are  the  same,  we  conclude  that 
the  consequents  are  proportional  (Prop.  IV.,  Cor.,  B.  II.)  ; 
therefore, 

Angle  ACD  :  angle  ACH  :  :  arc  AI  :  arc  AH. 

But  the  arc  AI  is  greater  than  the  arc  AH ;  therefore  the 
angle  ACD  is  greater  than  the  angle  ACH  (Def.  2,  B.  II.), 
that  is,  a  part  is  greater  than  the  whole,  which  is  absurd. 
Hence  the  angle  ACB  can  not  be  to  the  angle  ACD  as  the 
arc  AB  to  an  arc  greater  than  AD. 

In  the  same  manner,  it  may  be  proved  that  the  fourth  term 
of  the  proportion  can  not  be  less  than  AD  ;  therefore,  it  must 
be  AD,  and  we  ha\'e  the  proportion 

Angle  ACB     angle  A  ID  :  :  arc  AB  :  arc  AD. 
Cor.  1.  Since  the  an^le  at  the  center  of  a  circle,  and  the 


BOOK    III 


55 


arc  intercepted  by  its  sides,  are  so  related,  that  when  one  is 
increased  or  diminished,  the  other  is  increased  or  diminished 
in  the  same  ratio,  we  may  take  either  of  these  quantities  as 
the  measure  of  the  other.  Henceforth  we  shall  take  the  arc 
AB  to  measure  the  angle  ACB.  It  is  important  to  observe, 
that  in  the  comparison  of  angles,  the  arcs  which  measure 
them  must  be  described  with  equal  radii. 

Cor.  2.  In  equal  circles,  sectors  are  to  each  other  as  the  it 
arcs;  for  sectors  are  equal  when  their  angles  are  equal. 


PROPOSITION    XV.       THEOREM. 

An  inscribed  angle  is  measured  by  half  the  arc  included  be- 
tween its  sides. 

Let  BAD  be  an  angle  inscribed  in  the  circle  BAD.  The 
angle  BAD  is  measured  by  half  the  arc  BD. 

First.  Let  C,  the  center  of  the  circle, 
be  within  the  angle  BAD.  Draw  the  di- 
ameter AE,  also  the  radii  CB,  CD. 

Because  CA  is  equal  to  CB,  the  angle 
CAB  is  equal  to  the  angle  CBA  (Prop.  X., 
B.  I.) ;  therefore  the  angles  CAB,  CBA 
are  together  double  the  angle  CAB.  But 
the  angle  BCE  is  equal  (Prop.  XXVII.,  B. 
I.)  to  the  angles  CAB,  CBA ;  therefore, 
also,  the  angle  BCE  is  double  of  the  angle  BAG.  Now  the 
angle  BCE,  being  an  angle  at  the  center,  is  measured  by  the 
arc  BE :  hence  the  angle  BAE  is  measured  by  the  half  of 
BE.  For  the  same  reason,  the  angle  DAE  is  measured  by 
half  the  arc  DE.  Therefore,  the  whole  angle  BAD  is  meas- 
ured by  half  the  arc  BD. 

Second.  Let  C,  the  center  of  the  circle, 
be  without  the  angle  BAD.  Draw  the  di- 
ametei  AE.  It  may  be  demonstrated,  as 
in  the  first  case,  that  the  angle  BAE  is 
measured  by  half  the  arc  BE,  and  the  an- 
gle DAE  by  half  the  arc  DE  ;  hence  their 
difference,  BAD,  is  measured  by  half  of 
BD.  Therefore,  an  inscribed  angle,  &c. 

Cor.  1.  All  the  angles  BAG,  BDC,  &c., 
inscribed  in  the  same  segment  are  equal,  for  thev  are  all 
measured  by  half  the  same  arc  BEG.      (See  next  fig.) 

Cor.  2.  Every  angle  inscribed  in  a  semicircle  is  a  right 
angle,  because  it  is  measured  by  half  a  semicircumference 
th  it  is,  the  fourth  part  of  a  circumference 


56 


GEOMETRY. 


Cor.  3.  Every  angle  inscribed  in  a 
segment  greater  than  a  semicircle  is  an 
acute  angle,  for  it  is  measured  by  half 
an  arc  less  than  a  semicircumference. 

Every  angle  inscribed  in  a  segment 
less  than  a  semicircle  is  an  obtuse  an- 
gle, for  it  is  measured  by  half  an  arc 
greater  than  a  semicircumference. 

Cor.  4.  The  opposite  angles  of  an  in- 
scribed quadrilateral,  ABEC,  are  together  equal  to  two  right 
angles ;  for  the  angle  BAG  is  measured  by  half  the  arc  BEG, 
and  the  angle  BEG  is  measured  by  half  the  arc  BAG ;  there- 
fore the  two  angles  BAG,  BEG,  taken  together,  are  measured 
by  half  the  circumference  ;  hence  their  sum  is  equal  to  two 
right  angles. 


PROPOSITION  XVI.       THEOREM. 

The  angle  formed  by  a  tangent  and  a  chord,  is  measured  by 
half  the  arc  included  between  its  sides. 

Let  the  straight  line  BE  touch  the 
circumference  ACDF  in  the  point  A, 
and    from    A    let    the    chord    AC    be 
drawn  ;  the  angle  BAG  is  measured  by  _ 
half  the  arc  AFC. 

From  the  point  A  draw  the  diameter 
AD.  The  angle  BAD  is  a  right  angle 
(Prop.  IX.),  and  is  measured  by  half 
the  semicircumference  AFD  ;  also,  the  '  A 

angle  DAG  is  measured  by  half  the  arc  DC  (Prop.  XV.) ; 
therefore,  the  sum  of  the  angles  BAD,  DAC  is  measured  by 
half  the  entire  arc  AFDC. 

In  the  same  manner,  it  may  be  shown  that  the  angle  CAE 
is  measured  by  half  the  arc  AC,  included  between  its  sides. 

Cor.  The  angle  BAG  is  equal  to  an  angle  inscribed  in  the 
segment  AGC ;  and  the  angle  EAC  is  equal  to  an  angle  in 
scribed  in  the  segment  AFC. 


BOO!'     IV,  67 


BOOK  IV. 

THE  PROPORTIONS  CF  FIGURES. 
Definitions. 

1.  fiqual  figures  are  such  as  maybe  applied  the  one  to  tne 
other,  so  as  to  coincide  throughout.     Thus,  two  circles  having 
equal  radii  are  equal ;  and  two  triangles,  having  the  three  sides 
of  the  one  equal  to  the  three  sides  of  the  other,  each  to  each, 
are  also  equal. 

2.  Equivalent  figures   are   such  as  contain  equal  areas 
Two  figures  may  be  equivalent,  however  dissimilar.     Thus, 
a  circle  may  be  equivalent  to  a  square,  a  triangle  to  a  rec- 
tangle, &c. 

3.  Similar  figures  are  such  as  have  the  angles  of  the  one 
equal  to  the  angles  of  the  other,  each  to  each,  and  the  sides 
about  the  equal  angles  proportional.     Sides  which  have  the 
same  position  in  the  two  figures,  or  which  are  adjacent  to 
equal  angles,  are  called  homologous.     The  equal  angles  may 
also  be  called  homologous  angles. 

Equal  figures  are  always  similar,  but  similar  figures  may 
be  very  unequal.  k 

4.  Two  sides  of  one  figure  are  said  to  be  reciprocally  pro- 
portional to  two  sides  of  another,  when  one  side  of  the  first  is 
to  one  side  of  the  second,  as  the  remaining  side  of  the  sec- 
ond is  to  the  remaining  side  of  the  first. 

5.  In  different  circles,  similar  arcs,  sectors,  or  segments,  are 
those  which  correspond  to  equal  angles  at  the  center. 

Thus,  if  the  angles  A  and  D  are 
equal,  the  arc  BC  will  be  similar  to 
the  arc  EF,  the  sector  ABC  to  the 
sector  DEF,  and  the  segment  BGC 
to  th*.  segment  EHF.  B 

6.  The  altitude  of  a  triangle  is  the  perpen- 
dicular let  fall  from  the  vertex  of  an  angle 
on  the  opposite  side,  taken  as  a  base,  or  on 
the  base  produced 


58  GEOMETttY. 

7.  The  altitude  of  a  parallelogram  is  the 
perpendicular  drawn  to  the  base  from  the     > 
opposite  side.  Z. 

8.  The  altitude  of  a  trapezoid  is  the  distance 
between  its  parallel  sides. 


PROPOSITION  I.       THEOREM. 

Parallelograms  which  have  equal  bases  and  equal  altitudes 
are  equivalent. 

Let  the  parallelo- 
grams  ABCD,  ABEF 
Be  placed  so  that  their 
equal  bases  shall  coin- 
ride  with  each  other. 
Let  AB  be  the  common  A 

base  ;  and,  since  the  two  parallelograms  are  supposed  to  have 
the  same  altitude,  their  upper  bases,  DC,  FE,  will  be  in  the 
same  straight  line  parallel  to  AB. 

Now,  because  ABCD  is  a  parallelogram,  DC  is  equal  to 
A.B  (Prop.  XXIX.,  B.  I.).  For  the  same  reason,  FE  is  equal 
to  AB,  wherefore  DC  is  equal  to  FE ;  hence,  if  DC  arid  FE 
be  taken  away  from  the  same  line  DE,  the  remainders  CE 
and  DF  will  be  equal.  But  AD  is  also  equal  to  BC,  and  AF 
to  BE  ;  therefore  the  triangles  DAF,  CBE  are  mutually  equi 
lateral,  and  consequently  equal. 

Now  if  from  the  quadrilateral  ABED  we  take  the  triangle 
ADF,  there  will  remain  the  parallelogram  ABEF ;  and  if 
from  the  same  quadrilateral  we  take  the  triangle  BCE,  there 
will  remain  the  parallelogram  ABCD.  Therefore,  the  two 
parallelograms  ABCD,  ABEF,  which  have  the  same  base 
and  the  same  altitude,  are  equivalent. 

Cor.  Every  parallelogram  is  equivalent  to  the  rectangle 
which  has  the  same  base  and  the  same  altitude. 


PROPOSITION   II.       THEOREM. 

Every  triangle  is  half  of  the  parallelogram  which  has  the 
same  base  and  the  same  altitude. 

Let  the  parallelogram  ABDE  and  the  triangle  ABC  have 
the  same  base,  AB,  and  the  same  altitude ;  the  triangle  is 
half  of  the  parallelogram. 


BOOK    IV. 


Complete  the  parallelogram  ABFC  ;  c 
then  the  parallelogram  ABFC  is  equiv- 
alent to  the  parallelogram  ABDE,  be- 
cause they  have  the  same  base  and  the 
same  altitude  (Prop.  I.).  But  the  tri- 
angle  ABC  is  half  of  the  parallelogram 
ABFC  (Prop.  XXIX.,  Cor.,  B.  I.) ;  wherefore  the  triangle 
ABC  is  also  half  of  the  parallelogram  ABDE.  Therefore, 
every  triangle,  &c. 

Cor.  1.  Every  triangle  is  half  of  the  rectangle  which  has 
the  same  base  and  altitude. 

Cor.  2.  Triangles  which  have  equal  bases  and  equal  alti 
tudes  are  equivalent. 


PROPOSITION    III.       THEOREM. 


Two  rectangles  of  the  same  altitude,  are  to  each  other  as  their 
bases. 


•       j 

'       \       ' 

» 

I  i   i 

: 

! 

|       i       I 

I 

e 


13 


Let  ABCD,  AEFD  be  two  rec- 
tangles  which  have  the  common  al- 
titude AD  ;  they  are  to  each  other 
-s  their  bases  AB,  AE. 

Case  first.  When  the  bases  are  in 
the  ratio  of  two  whole  numbers,  for 
example,  as  7  to  4.  If  AB  be  divided  into  seven  equal  parts, 
AE  will  contain  four  of  those  parts.  At  each  point  of  divis- 
ion, erect  a  perpendicular  to  the  base  ;  seven  partial  rectan- 
gles will  thus  be  formed,  all  equal  to  each  other,  since  they 
have  equal  bases  and  altitudes  (Prop.  I.).  The  rectangle 
ABCD  will  contain  seven  partial  rectangles,  while  AEFD 
will  contain  four  ;  therefore  the  rectangle  ABCD%  is  to  the 
rectangle  AEFD  as  7  to  4,  or  as  AB  to  AE.  The  "same  rea- 
soning is  applicable  to  any  other  ratio  than  that  of  7  to  4 , 
therefore,  whenever  the  ratio  of  the  bases  can  be  expressed 
in  whole  numbers,  we  shall  have 

ABCD  :  AEFD  :  :  AB  :  AE. 

Case  second.  When  the  ratio  of  the  bases  can  not  be  ex- 
pressed in  whole  numbers,  it  is  still  true  that 
ABCD  :  AEFD  ::  AB    AE. 

For,  if  this  proportion  is  not  true,  the 

first  three  terms  remaining  the  same,  the 

fourth  must  be  greater  or  less  than  AE. 

Suppose  it  to  be  greater,  and  that  we  have 

ABCD  :  AEFD  :  :  AB  :  AG. 

Conceive  the  line  AB  to  be  divided  into 


FI 


EHG-    B 


60 


rfDMETRY. 


D 


equal  parts,  each  less  than  EG ;  there  will 
be  at  least  one  point  of  division  between 
E  and  G.  Let  H  be  that  point,  and  draw 
the  perpendicular  HI.  The  bases  AB,  AH 
will  be  to  each  other  in  the  ratio  of  two 
whole  numbers,  and  by  the  preceding  case 
we  shall  have 

ABCD  :  AHID  :  :  AB  :  AH. 

But,  by  hypothesis,  we  have 

ABCD  :  AEFD  :  :  AB  :  AG. 

In  these  two  proportions  the  antecedents  are  equal ;  thei  e- 
fore  the  consequents  are  proportional  (Prop.  IV.,  Cor.,  B.  II.), 
and  we  have 

AHID  :  AEFD  :  :  AH  :  AG. 

But  AG  is  greater  than  AH ;  therefore  the  rectangle 
AEFD  is  greater  than  AHID  (Def.  2,  B.  II.)  ;  that  is,  a  part  is 
greater  than  the  whole,  which  is  absurd.  Therefore  ABCD 
can  not  be  to  AEFD  as  AB  to  a  line  greater  than  AE. 

In  the  same  manner,  it  may  be  shown  that  the  fourth  term 
of  the  proportion  can  not  be  less  than  AE ;  hence  it  must  be 
AE,  and  we  have  the  proportion 

ABCD  :  AEFD  :  :  AB  :  AE. 
Therefore,  two  rectangles,  &c. 


PROPOSITION  IV.       THEOREM. 

Any  two  rectangles  are  to  each  other  as  the  products  of  then 
bases  by  their  altitudes. 

Let  ABCD,  AEGF  be  two  rectangles ;  the  ratio  of  the  rec- 
tangle ABCD  to  the  rectangle  AEGF,  is  the  same  with  the 
ratio  of  the  product  of  AB  by  AD,  to  ths  product  of  AE  by 
AF ;  that  is, 

ABCD  :  AEGF  : :  AB  X  AD  :  AE  x  AF. 

Having  placed  the  two  rectangles  so 
that  the  angles  at  A  are  vertical,  pro- 
duce the  sides  GE,  CD  till  they  meet  in 
H.  The  two  rectangles  ABCD,  AEHD 
have  the  same  altitude  AD  ;  they  are, 
therefore,  as  their  bases  AB,  AE  (Prop. 
III.).  So,  also,  the  rectangles  AEHD,  . 
AEGF,  having  the  same  altitude  AE, 
are  to  each  other  as  their  bases  AD,  AF 
two  proportions 

ABCD  :  AEHD  :  :  AB    AE, 
AEHD  :  AEGF  :  :  AD    AF. 


II 


F 

Thus,  we  have  the 


BOOK    IV  61 

Hence  (Prop.  XL,  Cor.,  B.  II.), 

ABCD  :  AEGF  :  :  ABxAD  :  AExAF. 

Scholium.  Hence  we  may  take  as  the  measure  of  a  rec- 
tangle the  product  of  its  base  by  its  altitude ;  provided  we  un- 
derstand by  it  the  product  of  two  numbers,  one  of  which  ia 
the  number  of  linear  units  contained  in  the  base,  and  the  oth- 
er the  number  of  linear  units  contained  in  the  altitude. 


PROPOSITION  V.       THEOREM. 

The  area  of  a  parallelogram  is  equal  to  the  product  of  its 
base  by  its  altitude. 

Let  ABCD  be  a  parallelogram,  AF  its  -p  D  EC 
altitude,  and  AB  its  base ;  then  is  its  sur- 
face measured  by  the  product  of  AB  by 
AF.  For,  upon  the  base  AB,  construct  a 
rectangle  having  the  altitude  AF  ;  the  par- 
allelogram ABCD  is  equivalent  to  the  rec- 
tangle ABEF  (Prop.  L,  Cor.).  But  the  rectangle  ABEF  is 
measured  by  AB  X  AF  (Prop.  IV.,  Schol.)  ;  therefore  the  area 
of  the  parallelogram  ABCD  is  equal  to  AB  X  AF. 

Cor.  Parallelograms  of  the  same  base  are  to  each  other  as 
their  altitudes,  and  parallelograms  of  the  same  altitude  are 
to  each  other  as  their  bases ;  for  magnitudes  have  the  same 
ratio  that  their  equimultiples  have  (Prop.  VIIL,  B.  II.). 


PROPOSITION    VI.       THEOREM. 

The  area  of  a  triangle  is  equal  to  half  the  product  of  its 
base  by  its  altitude. 

Let  ABC  be  any  triang.e,  BC  its  base,  and 
AD  its  altitude  ;  the  area  of  the  triangle  ABC 
«i  measured  by  half  the  product  of  BC  by  AD. 

For,  complete  the  parallelogram  ABCE. 
The  triangle  ABC  is  half  of  the  parallelo- 
gram  ABCE  (Prop.  II.)  ;  but  the  area  of  the  J 
parallelogram  is  equal  to  BC  x  AD  (Prop.  V.)  ;  hence  the 
area  of  the  triangle  is  equal  to  one  half  of  the  product  of 
BC  by  AD.  Therefore,  the  area  of  a  triangle,  &c. 

Cor.  1.  Triangles  of  the  same  altitude  are  to  each  other 
as  their  bases,  and  triangles  of  the  same  base  are  to  each  oth- 
er as  their  altitudes. 

Cor  2    Equivalent  triangles,  whose  bases  are  equal,  have 


GEOMETliy. 


equal  altituaes  ;  and  equivalent  triangles,  whose  altitudes  are 
equal,  have  equi.l  bases. 


PROPOSITION    VII.       THEOREM. 

The  area  of  a  trapezoid  is  equal  to  half  the  product  of  its 
altitude  by  the  sum  of  its  parallel  sides. 

Let  ABCD  be  a  trapezoid,  DE  its  al- 
titude, AB  and  CD  its  parallel  sides ; 
'ts  area  is  measured  by  half  the  product 
of  DE,  by  the  sum  of  its  sides  AB,  CD. 

Bisect  BC  in  F,  and  through  F  draw 
GH  parallel  to  AD,  and  produce  DC  to  A  E  & B 

H.  In  the  two  triangles  BFG,  CFH, 
the  side  BF  is  equal  to  CF  by  construction,  the  vertical  an- 
gles BFG,  CFH  are  equal  (Prop.  V.,  B.  I.),  and  the  angle 
FCH  is  equal  to  the  alternate  angle  FBG,  because  CH  and 
BG  are  parallel  (Prop.  XXIIL,  B.  I.) ;  therefore  the  triangle 
CFH  is  equal  to  the  triangle  BFG.  Now,  if  from  the  whole 
figure,  ABFHD,  we  take  away  the  triangle  CFH,  there  will 
remain  the  trapezoid  ABCD ;  and  if  from  the  same  figure, 
ABFHD,  we  take  away  the  equal  triangle  BFG,  there  will 
'•emain  the  parallelogram  AGHD.  Therefore  the  trapezoid 
ABCD  is  equivalent  to  the  parallelogram  AGHD,  and  is 
measured  by  the  product  of  AG  by  DE. 

Also,  because  AG  is  equal  to  DH,  and  BG  to  CH,  there- 
rore  the  sum  of  AB  and  CD  is  equal  to  the  sum  of  AG  and 
DH,  or  twice  AG.  Hence  AG  is  equal  to  half  the  sum  of 
the  parallel  sides  AB,  CD ;  therefore  the  area  of  the  trape- 
zoid ABCD  is  equal  to  half  the  product  of  the  altitude  DE 
by  the  sum  of  the  bases  AB,  CD. 

Cor.  If  through  the  point  F,  the  middle  of  BC,  we  draw 
FK  parallel  to  the  base  AB,  the  point  K  will  also  be  the  mid- 
dle of  AD.  For  the  figure  AKFG  is  a  parallelogram,  as 
also  DKFH,  the  opposite  sides  being  parallel.  Therefore 
AK  is  equal  to  FG,  and  DK  to  HF.  But  FG  is  equal  to  FH, 
since  the  triangles  BFG,  CFH  are  equal ;  therefore  AK  is 
equal  to  DK. 

Now,  since  KF  is  equal  to  AG,  the  area  of  the  trapezoid  is 
equal  to  DE  xKF.  Hence  the  area  of  a  trapezoid  is  equal  to 
its  altitude,multipliea  by  the  line  which  joins  the  middle  points 
of  the  sides  u  hi  'k  are  not  parallel. 


BOOK    IV.  63 


PROPOSITION  VIII.       THEOREM. 

If  a  straight  line  is  divided  into  any  two  parts,  the  square  oj 
the  whole  line  is  equivalent  to  the  squares  of  the  two  parts,  to- 
gether with  twice  the  rectangle  contained  by  \he  parts. 

Let  the  straight  line  AB  be  divided  into  any  two  parts  in 
C ;  the  square  on  AB  is  equivalent  to  the  squares  on  AC 
CB,  together  with  twice  the  rectangle  contained  by  AC,  CB 
that  is, 

AB%  or  (AC+CB)a=AC2+CB2+2ACxCB. 

Upon  AB  describe  the   square   ABDE ;     E  j[     D 

take  AF  equal  to  AC,  through  F  draw  FG 
parallel  to  AB,  and  through  C  draw  CH  par- 
allel  to  AE. 

The   square  ABDE  is  divided  into  four 
parts :  the  first,  ACIF,  is  the  square  on  AC, 


since  AF  was  taken  equal  to  AC.  The  sec-  -A- 
ond  part,  IGDH,  is  the  square  on  CB ;  'for,  because  AB  is 
equal  to  AE,  and  AC  to  AF,  therefore  BC  is  equal  to  EF 
(Axiom  3,  B.  I.).  But,  because  BCIG  is  a  parallelogram, 
GI  is  equal  to  BC ;  and  because  DEFG  is  a  parallelogram, 
DG  is  equal  to  EF  (Prop.  XXIX.,  B.  I.) ;  therefore  HIGD  is 
equal  to  a  square  described  on  BC.  If  these  two  parts  are 
taken  from  the  entire  square,  there  will  remain  the  two  rect- 
angles BCIG,  EFIH,  each  of  which  is  measured  by  AC  X 
OB ;  therefore  the  whole  square  on  AB  is  equivalent  to  the 
squares  on  AC  and  CB,  together  with  twice  the  rectangle  of 
AC  x  CB.  Therefore,  if  a  straight  line,  &c. 

Cor.  The  square  of  any  line  is  equivalent  to  four  times  the 
square  of  half  that  line.  For,  if  AC  is  equal  to  CB,  the  four 
figures  AI,  CG,  FH,  ID  become  equal  squares. 

Scholium.  This  proposition  is  expressed  algebraicaJv 
thus: 


PROPOSITION    IX.       THEOREM. 

The  square  described  on  the  difference  of  two  lines,  is  equiv 
alent  to  the  sum  of  the  squares  of  the  lines,  diminished  by  twice 
the  rectangle  contained  by  the  lines. 

Let  AB,  BC  be  an>  two  lines,  and  AC  their  difference ; 
the  square  described  on  AC  is  equivalent  to  the  sum  of  the 


GEOMETRY. 


I   E 


C      B 


squares  on  AB  and  CB,  diminished  by  twice  the  rectangle 
contained  by  AB,  CB ;  that  is, 

AC2,  or  (AB  — BC)2=AB2+BC2— SABxBC. 

Upon  AB  describe  the  square  ABKF;  ^     p  G     K 

take  AE  equal  to  AC,  through  C  draw 
CG  parallel  to  BK,  and  through  E  draw 
HI  parallel  to  AB,  and  complete  the 
square  EFLI. 

Because  AB  is  equa".  to  AF,  and  AC  to 
AE  ;  therefore  CB  is  equal  to  EF,  and  GK 
to  LF.  Therefore  LG  is  equal  to  FK  or  AB ;  and  hence  the 
two  rectangles  CBKG,  GLID  are  each  measured  by  AB  X 
BC.  If  these  rectangles  are  taken  from  the  entire  figure 
ABKLIE,  which  is  equivalent  to  AB2-f  BC2,  there  will  evi- 
dently remain  the  square  ACDE.  Therefore,  the  square 
described,  &c. 

Scholium.     This    proposition    is    expressed    algebraically 
hus: 

(a — b)9=ai — 2ab+b\ 

Cor.  (a+b)*—(a—by=4ab. 


PROPOSITION    X.       THEOREM. 


H 


Fhe  rectangle  contained  by  the  sum  and  difference  of  two 
lines,  is  equivalent  to  the  difference  of  the  squares  of  those  lines 

Let  AB,  BC  be  any  two  lines  ;  the  rectangle  contained  by 
the  sum  and  difference  of  AB  and  BC,  is  equivalent  to  the 
difference  of  the  squares  on  AB  and  BC ;  that  is, 
(AB+BC)  X  (AB  — BC)  =AB2  — BC2. 

Upon  AB  describe  the  square  ABKF,    3?          &     K 
and  upon  AC  describe  the  square  ACDE ; 
produce  AB  so  that  BI  shall  be  equal  to 
BC,  and  complete  the  rectangle  AILE. 

The  base  AI  of  the  rectangle  AILE  is 
the  sum  of  the  two  lines  AB,  BC,  and  its 
altitude  AE  is  the  difference  of  the  same  -^  C  B  I 

lines ;  therefore  AILE  is  the  rectangle  contained  by  the  sum 
and  difference  of  the  lines  AB,  BC.  But  this  rectangle  is 
composed  of  the  two  parts  ABHE  and  BILH  ;  and  the  part 
BILH  is  equal  to  the  rectangle  EDGF,  for  BH  is  equal  to 
DE,  and  BI  is  equal  to  EF.  Therefore  AILE  is  equivalent 
to  the  figure  ABHDGF.  But  ABHDGF  is  the  excess  of  the 
square  ABKF  above  the  square  DHKG,  which  is  the  square 
of  BC ;  therefore, 

<AB-fBC)x(AB— BC)=AB2  —  BC2. 


BOOK    IV. 


Scholium.     This   proposition   is    expressed    algebraical!^ 
thus: 

i<— 6=aa— b*. 


L    E 


PROPOSITION  XI.       THEOREM. 

In  any  right-angled  triangle,  the  square  described  on  the  ny> 
fothenuse  is  equivalent  to  the  sum  of  the  squares  on  the  other 
twc  sides. 

Let  ABC  be  a  right-angled  triangle, 
having    the    right    angle    BAG ;    the 
square  described  upon  the  side  BC  is  _ 
equivalent  to  the  sum  of  the  squares 
upon  BA,  AC. 

On  BC  describe  tne  square  BCED, 
and  on  BA,  AC  the  squares  BG,  CH ; 
and  through  A  draw  AL  parallel  to 
BD,  and  join  AD,  FC. 

Then,  because   each   of  the  angles 
BAG,  BAG  is  a  right  angle,  CA  is  in 
the  same  straight  line  with  AG  (Prop.  III.,  B.  I.).     For  the 
same  reason,  BA  and  AH  are  in  the  same  straight  line. 

The  angle  ABD  is  composed  of  the  angle  ABC  and  the 
right  angle  CBD.  The  angle  FBC  is  composed  of  the  same 
angle  ABC  and  the  right  angle  ABF;  therefore  the  whole 
angle  ABD  s  equal  to  the  angle  FBC.  But  AB  is  equal  to 
BF,  being  sides  of  the  same  square ;  and  BD  is  equal  to  BC 
for  the  same  reason  ;  therefore  the  triangles  ABD,  FBC  have 
two  sides  and  the  included  angle  equal ;  they  are  therefore 
equal  (Prop.  VI.,  E.I.). 

But  the  rectangle  BDLK  is  double  of  the  triangle  ABDS 
because  they  have  the  same  base,  BD,  and  the  same  altitude, 
BK  (Prop.  II.,  Cor.  1) ;  and  the  square  AF  is  double  of  the 
triangle  FBC,  for  they  have  the  same  base,  BF,  and  the  same 
altitude,  AB.  Now  the  doubles  of  equals  are  equal  to  one 
another  (Axiom  6,  B.  I.) ;  therefore  the  rectangle  BDLK  is 
equivalent  to  the  square  AF. 

In  the  same  manner,  it  may  be  demonstrated  that  the  rec- 
tangle CELK  is  equivalent  to  the  square  AI ;  therefore  the 
whole  square  BCED,  described  on  the  hypothenuse,  is  equiv- 
alent to  the  two  squares  ABFG,  ACIH,  described  on  the  two 
other  sides  ;  that  is, 

BCa=AB'4-ACa. 

Cor.  1.  The  square  of  one  of  the  sides  of  a  right-angled 


0(5  GEOMETRY, 

triangle  is  equivalent  to  the  square  of  the  hypothemise,  dimin 
ished  bj  the  square  of  the  other  side  ;  that  is, 

AB2=BC2— AC2. 

Cor.  2.  The  square  BCED,  and  the  rectangle  BKLD,  hav- 
ing the  same  altitude,  are  to  each  other  as  their  bases  BC, 
BK  (Prop.  III.).  But  the  rectangle  BKLD  is  equivalent  to 
the  square  AF ;  therefore, 

BC3  :  AB3 :  :  BC  :  BK. 
In  the  same  manner, 

BC3  :  AC3  :  :  BC  :  KC. 
Therefore  (Prop.  IV.,  Cor.,  B.  II.), 

AB3  :  AC2  :  :  BK  :  KC. 

That  is,  in  any  right-angled  triangle,  if  a  line  be  drawn 
from  the  right  angle  perpendicular  to  the  hypothenuse,  the 
squares  of  the  two  sides  are  proportional  to  the  adjacent  seg- 
ments of  the  hypothenuse  ;  also,  the  square  of  the  hypothenuse 
is  to  the  square  of  either  of  the  sides,  as  the  hypothenuse  is  to 
the  segment  adjacent  to  that  side. 

Cor.  3.  Let  ABCD  be  a  square,  and  AC  its 
diagonal ;  the  triangle  ABC  being  right-angled 
and  isosceles,  we  have 

AC2=AB2+BC2=2ABa; 

therefore  the  square  described  on  the  diagonal  of  a 
square,  is  double  of  the  square  described  on  a  side. 
If  we  extract  the  square  root  of  each  mem- 
ber of  this  equation,  we  shall  have 

AC=ABv/2 ;  or  AC  :  AB  :  :  ^/2  :  1. 


PROPOSITION   XII.       THEOREM. 

In  any  triangle,  the  square  of  a  side  opposite  an  acute  angle, 
is  less  than  the  squares  of  the  base  and,  of  the  other  side,  by 
twice  th-3  rectangle  contained  by  the  base,  and  the  distance  from 
the  acute  angle  to  the  foot  of  the  perpendicular  let  fall  from  the 
opposite  angle. 

Let  ABC  be  any  triangle,  and  the  angle  at  C  one  of  its 
acute  angles,  and  upon  BC  let  fall  the  perpendicular  AD  from 
the  opposite  angle  ;  then  will 

AB2=BCa-rAC3— -SBCxCD. 

First.  When  the  perpendicular  falls  with- 
in the  triangle  ABC,  we  have  BD  =  BC-  CD, 
and  therefore  BD2=BC2-I-CD2— 2BC  xCD 
(Prop.  IX.).  To  each  of  these  equals  add 
AD2;  then  BD2+AD2  =  BC2+CD2  f  AD2— 
SBC  X  CD  But  ii.  trie  right- angled  triangle  B* 


IV. 


A 


ABD.  BD'+AD2=AB2  ;  and  in  the  triangle   ADC.  CD 
AD*=AC2  (Prop.  XL);  therefore 

AB2=BC2-f  AC2— 2BC  xCD. 

Secondly.  When  the  perpendicular  falls 
without  the  triangle  ABC,  we  have  BD  = 
CD— BC,  and  therefore  BD2=CD3-fBC2— 
2CDXBC  (Prop.  IX.).  To  each  of  these 
equals  add  AD2 ;  then  BD2+AD2=CD2+AD2 
+BC2— 2CDXBC.  But  BD2+AD2=AB2  ; 
andCD2+AD2=AC2;  therefore 

AB2=BC2+AC2— 2EC  xCD. 

Scholium.     When  the  perpendicular  AD  falls  upon  AB, 
this  proposition  reduces  to  the  same  as  Prop.  XL,  Cor.  1. 


D     B 


G 


PROPOSITION    XIII.       THEOREM. 

In  obtuse-angled  triangles,  the  square  of  the  side  opposite 
i\e  obtuse  angle,  is  greater  than  the  squares  of  the  base  and  the 
ether  side,  by  twice  the  rectangle  contained  by  the  base,  and  the 
distance  from  the  obtuse  angle  to  the  foot  of  the  perpendicular 
let  fall  from  the  opposite  angle  on  the  base  produced. 

Let  ABC  be  an  obtuse-angled  triangle,  having  the  obtuse 
angle  ABC,  and  from  the  point  A  let  AD  be  drawn  perpen- 
dicular to  BC  produced ;  the  square  of  AC  is  greater  than 
the  squares  of  AB,  BC  by  twice  the  rectangle  BC  X  BD. 

For  CD  is  equal  to  BC+BD  ;  therefore  CD2 
r-BC2-J-BD2+2BCxBD  (Prop.  VIII.).  To 
^ach  of  these  equals  add  AD2 ;  then  CD2+ 
AD2  -  BC2+BD2+ AD2+2BC  X  BD.  But  AC2 
is  equal  to  CD2+AD2  (Prop.  XL),  and  AB2  is 
equal  to  BD2-fAD2 ;  therefore  AC2=BC2+ 
AB2+2BCxBD.  Therefore,  in  obtuse-an- 
gled triangles,  &c. 

Scholium.  The  right-angled  triangle  is  the  only  one  in 
which  the  sum  of  the  squares  of  two  sides  is  equivalent  to  the 
square  on  the  third  side ;  for,  if  the  angle  contained  by  the 
two  sides  is  acute,  the  sum  of  their  squares  is  greater  than 
the  square  of  the  opposite  side ;  if  obtuse,  it  is  less. 


D     B 


PROPOSITION   XIV.       THEOREM. 


In  any  triangle,  if  a  straight  line  is  drawn  from  the 
to  the  middle  of  the  base,  the  sum  of  the  squares  of  the  other  two 
sides  is  equivalent  to  twice  the  square  of  the  bisecting  line,  to- 
gether with  twice  the  square  of  half  the  base. 

Let  ABC  bo  a  trangte  having  a  line  AD  drawn  from  the 


68  GEOMETRY. 

middle  of  the  base  to  the  opposite  angle ;  the  squares  of 
and  AC  are  together  double  of  the  squares  of  AD  and  BP 

From  A  draw  AE  perpendicular  to  BC ; 
then,  in  the  triangle  ABD,  by  Prop.  XIIL, 

AB2=AD8+DB2+2DBxDE; 
and,  in  the  triangle  ADC,  by  Prop.  XII., 

AC2= AD'+DC"  — 2DC  x  DE. 
Hence,  by  adding  these  equals,  and  ob- 
serving that  BD=DC,  and  therefore  BD2  =  B 
DC2,  and  DBxDE=DCxDE,  we  obtain 

AB2+AC2=2AD2+2DB2. 
Therefore,  in  any  triangle,  &c. 


PROPOSITION    XV.       THEOREM. 

In  every  parallelogram  the  squares  of  the  sides  are  togetfic* 
equivalent  to  the  squares  of  the  diagonals. 

Let  ABCD  be  a  parallelogram,  of  which  A.  D 

the  diagonals  are  AC  and  BD ;  the  sum  of  ^~~ 
the  squares  of  AC  and  BD  is  equivalent  to 
the  sum  of  the  squares  of  AB,  BC,  CD,  DA. 

The  diagonals  AC  and  BD  bisect  each       B 
other  in  E  (Prop.  XXXIL,  B~  I.)  ;  therefore,  in  the  triangle 
ABD  (Prop.  XIV.), 

ABM-AD»=2BE2+2AE2 ; 
and,  in  the  triangle  BDC, 

CD2+BC2=2BE2+2EC2. 

Adding  these  equals,  and  observing  that  AE  is  equal  to 
EC,  we  have 

AB2+BC2+CD2+AD2=4BE2+4AE2. 
But  4BE9=BD2,and  4AE2=ACa  (Prop.  VIIL,  Cor.)  ;  there- 
fore 

AB2+BCa+CD3+AD2=BD2+AC2. 
Therefore,  in  every  parallelogram,  &c. 


PROPOSITION  XVI.       THEOREM. 


If  a  straight  line  be  drawn  parallel  to  the  base  of  a  triangfe, 
it  will  cut  the  other  sides  proportionally ;  and  if  the  sides  be 
cut  proportionally,  the  cutting  line  will  be  parallel  to  the  base 
of  the  triangle. 

Let  DE  be  drawn  parallel  to  BC,  the  base  of  the  triangle 
ABC  then  will  AD  DB  : :  AE  :  EC. 


BCOK    IV. 


D 


Join  BE  and  DC  ;  then  the  triangle  BDE  is 
equivalent  to  the  triangle  DEC,  because  they 
nave  the  same  base,  DE,  and  the  same  altitude, 
since  their  vertices  B  and  C  are  in  a  line  par- 
allel to  the  base  (Prop.  II.,  Cor.  2). 

The  triangles  ADE,  BDE,  whose  common 

vertex  is  E,  having  the  same  altitude,  are  to     ^___ 

each  other  as  their  bases  AD,  DB  (Prop.  VI.,  B  C 

Cor.  1)  ;  hence 

ADE  :  BDE  :  :  AD  :  DB. 

The  triangles  ADE,  DEC,  whose  common  vertex  is  D, 
having  the  same  altitude,  are  to  each  other  as  their  bases 
AE,  EC ;  therefore 

ADE  :  DEC  :  :  AE  :  EC. 

But,  since  the  triangle  BDE  is  equivalent  to  the  triangle 
DEC,  therefore  (Prop.  IV.,  B.  II.), 

AD  :  DB  :  :  AE  :  EC. 

Conversely,  let  DE  cut  the  sides  AB,  AC,  so  that  AD  :  DB 
:  :  AE  :  EC ;  then  DE  will  be  parallel  to  BC. 

For  AD  :  DB  :  :  ADE  :  BDE  (Prop.  VI.,  Cor.  1) ;  and  AE 
:  EC  :  :  ADE  :  DEC ;  therefore  (Prop.  IV.,  B.  II.),  ADE  : 
BDE  :  :  ADE  :  DEC ;  that  is,  the  triangles  BDE,  DEC  have 
the  same  ratio  to  the  triangle  ADE ;  consequently,  the  trian- 
gles BDE,  DEC  are  equivalent,  and  having  the  same  base  DE, 
their  altitudes  are  equal  (Prop.  VI.,  Cor.  2),  that  is,  they  are 
between  the  same  parallels.  Therefore,  if  a  straight  line,  &c. 

Cor.  1.  Since,  by  this  proposition,  AD  :  DB  :  :  AE  :  EC ; 
by  composition,  AD+DB  :  AD  :  :  AE+EC  :  AE  (Prop.  VI., 
B.  II.),  or  AB  :  AD  :  :  AC  :  AE  ;  also,  AB  :  BD  :  :  AC  :  EC. 

Cor.  2.  If  two  lines  be  drawn  parallel  to  the 
base  of  a  triangle,  they  will  divide  the  other  sides 
proportionally.  For,  because  FG  is  draw** 
parallel  to  BC,  by  the  preceding  proposition, 
AF  :  FB  :  :  AG  :  GC.  Also,  by  the  last  cor- 
ullary,  because  DE  is  parallel  to  FG,  AF  :  DF 
.  :  AG  :  EG.  Therefore  DF  :  FB  :  :  EG  :  GC 
(Prop.  IV.,  Cor.,  B.  II.).  Also,  AD  :  DF  :  :  B  C 

AE  :  EG. 

Cor.  3.  If  any  number  of  lines  be  drawn  parallel  to  the 
base  of  a  triangle,  the  sides  will  be  cut  proportionally. 


PROPOSITION    XVII.       THEOREM. 


The  line  which  bisects  the  vertical  angle  of  a  triangle,  di- 
vides the  base  into  two  segments,  which  are  proportional  to  the 
adjacent  *ide*. 


GEOMETRY. 


Let  the  angle  BAG  of  the  triangle  ABC  be  bisected  by  th* 
straight  line  AD  ;  then  will  a 

BD  :  DC  :  :  BA  :  AC. 

Through  the  point  B  draw  BE  par- 
allel to  DA,  meeting  CA  produced  in  E. 
The  triangle  ABE  is  isosceles.  For, 
since  AD  is  parallel  to  EB,  the  angle 
ABE  is  equal  to  the  alternate  angle 
DAB  (Prop.  XXIII.,  B.  I.)  ;  and  the  exterior  angle  CAD  is 
equal  to  the  interior  and  opposite  angle  AEB.  But,  by  hy- 
pothesis, the  angle  DAB  is  equal  to  the  angle  DAC ;  there- 
fore the  angle  ABE  is  equal  to  AEB,  and  the  side  AE  to  the 
side  AB  (Prop.  XL,  B.  I.). 

And  because  AD  is  drawn  parallel  to  BE,  the  base  of  the 
triangle  BCE  (Prop.  XVI.), 

BD  :  DC  :  :  EA  :  AC. 

But  AE  is  equal  to  AB,  therefore 

BD  :  DC  :  :  BA  :  AC. 

Therefore,  the  line,  &c. 

Scholium.  The  line  which  bisects  the  exterior  angle  ol  a 
triangle,  divides  the  base  produced  into  segments,  which  are 
proportional  to  the  adjacent  sides. 

Let  the  line  AD  bisect  the  exterior  . 

angle  C  AE  of  the  triangle  ABC  ;  then 


BD  :  DC  :  :  BA  :  AC. 

Through  C  draw  CF  parallel  to 
AD ;  then  it  may  be  proved,  as  in  the  .. 
preceding  proposition,  that  the  angle 

ACF  is  equal  to  the  angle  AFC,  and  AF  equal  to  AC.     And 
because  FC  is  parallel  to  AD  (Prop.  XVI.,  Cor.  1),  BD  :  DO 
:  BA  :  AF.  But  AF  is  equal  to  AC ;  therefore 
BD  :  DC  :  :  BA  :  AC. 


PROPOSITION    XVIII.       THEOREM. 

Equiangular  triangles  have  their  homologous  sides  propor* 
fional,  and  are  similar. 

Let  ABC,  DCE  be  two  equiangular 
triangles,  having  the  angle  BAG  equal  to 
the  angle  CDE,  and  the  angle  ABC  equal 
to  the  angle  DCE,  and,  consequently,  the 
angle  ACB  equal  to  the  angle  DEC  ;  then 
the  homologous  sides  will  be  proportion- 
al, and  we  shall  have 

BC  :  CE  :  :  BA  :  CD  :  :  AC  :  DE. 


BOCK    IV. 


71 


Place  the  triangle  DCE  so  that  the  side  CE  may  be  COD 
tiguous  to  BC,  and  in  the  same  straight  line  with  it ;  and  pro- 
duce the  sides  BA,  ED  till  they  meet  in  F. 

Because  BCE  is  a  straight  line,  and  the  angle  ACB  is 
equal  to  the  angle  DEC,  AC  is  parallel  to  EF  (Prop.  XXII., 
B.  L).  Again,  because  the  angle  ABC  is  equal  to  the  angle 
DCE,  the  line  AB  is  parallel  to  DC ;  therefore  the  figure 
A  CDF  is  a  parallelogram,  and,  consequently,  AF  is  equal  to 
CD,  and  AC  to  FD  (Prop.  XXIX.,  B.  L). 

And  because  AC  is  parallel  to  FE,  one  of  the  sides  of  the 
triangle  FBE,  BC  :  CE  :  :  BA  :  AF  (Prop.  XVI.)  ;  but  AF  is 
equal  to  CD  ;  therefore 

BC  :  CE  :  :  BA  :  CD. 

Again,  because  CD  is  parallel  to  BF,  BC  :  CE  :  :  FD  :  DE 
But  FD  is  equal  to  AC  ;  therefore 

BC  :  CE  :  :  AC  :  DE. 

And,  since  these  two  proportions  contain  the  same  ratio 
BC  :  CE,  we  conclude  (Prop.  IV.,  B.  II.) 
BA  :  CD  :  :  AC  :  DE. 

Therefore  the  equiangular  triangles  ABC,  DCE  have  then 
homologous  sides  proportional ;  hence,  by  Def.  3,  they  are 
similar. 

Cor.  Two  triangles  are  similar  when  they  have  two  an 
gles  equal,  each  to  each,  for  then  the  third  angles  must  also 
be  equal. 

Scholium.  In  similar  triangles  the  homologous  sides  are 
opposite  to  the  equal  angles;  thus,  the  angle  ACB  being 
equal  to  the  angle  DEC,  the  side  AB  is  homologous  to  DC, 
and  so  with  the  other  sides. 


PROPOSITION   XIX.       THEOREM. 

Two  triangles  which  have  their  homologous  sides  proportion- 
al, are  equiangular  and  similar. 

Let  the  triangles  ABC,  DEF 
have  their  sides  proportional,  so 
that  BC  :  EF  :  :  AB  :  DE  :  :  AC 
:  DF ;  then  will  the  triangles 
have  their  angles  equal,  viz. : 
the  angle  A  equal  to  the  angle 
D,  B  equal  to  E,  and  C  equal  to  I 
F. 

At  the  point  E,  in  the  straight 
line  EF,  make  the  angle  FEG  equal  to  B,  and  at  tiie  point  f 
make  the  angle  EFG  equal  to  C  ;  the  third  angle  G  will  b« 


72  GEOMETRY 

equa.  to  the  third  angle  A,  and  the  vwo  triangles  ABC,  GEF 
will  be  equiangular  (Prop.  XXVIL,  Cor.  2,  B.  I.)  ;  therefore 
6y  the  preceding  theorem, 

BC  :  EF  :  :  AB  :  GE. 
But,  by  hypothesis, 

BC  :  EF  :  :  AB  :  DE ; 
therefore  GE  is  equal  to  DE. 
Also,  by  the  preceding  theorem, 

BC:EF::  AC:GF; 
but,  by  hypothesis, 

BC  :  EF  :  :  AC  :  DF ; 

consequently,  GF  is  equal  to  DF.  Therefore  the  triangles 
GEF,  DEF  have  their  three  sides  equal,  each  to  each  ;  hence 
their  angles  also  are  equal  (Prop.  XV.,  B,  I.).  But,  by  con- 
struction, the  iriangle  GEF  is  equiangular  to  the  triangle 
ABC  ;  therefore,  also,  the  triangles  DEF,  ABC  are  equiangu- 
lar and  similar.  Wherefore,  two  triangles,  &c. 


PROPOSITION  XX.       THEOREM. 

Two  triangles  are  similar,  when  they  have  an  angle  of  the 
one  equal  to  an  angle  of  the  other,  and  the  sides  containing 
those  angles  proportional. 

Let  the  triangles  ABC,  DEF  have  the  angle  A  of  the  one, 
equal  to  the  angle  D  of  the  other,  and  let  AB  :  DE  :  :  AC 
DF ;  the  triangle  ABC  is  similar  to  the  triangle  DEF. 

Take  AG  equal  to  DE,  also  AH  A. 
equal  to  DF,  and  join  GH.  Then 
the  triangles  AGH,  DEF  are  equal, 
since  two  sides  and  the  included 
angle  in  the  one,  are  respectively 
equal  to  two  sides  and  the  included 

angle  in  the  other  (Prop.  VI.,  B.  I.).     

But,  by  hypothesis,  AB  :  DE  :  :  AC  B  C    E 

.  DF ;  therefore 

AB  :  AG  :  :  AC  :  AH ; 

that  is,  the  sides  AB,  AC,  of  the  triangle  ABC,  are  cut  pro- 
portionally by  the  line  GH ;  therefore  GH  is  parallel  to  BC 
(Prop.  XVI.).  Hence  (Prop.  XXIII.,  B.  I.)  the  angle  AGH  is 
equal  to  ABC,  and  the  triangle  AGH  is  similar  to  the  trian- 
gle ABC.  But  the  triangle  DEF  has  been  shown  to  be 
equal  to  the  triangle  AGH ;  hence  the  triangle  DEF  is  simi- 
'ar  to  the  triangle  ABC.  Therefore;  two  triangles,  &c. 


BOOK    IV.  73 


PROPOSITION    XXI.       THEOREM. 

Two  triangles  are  similar,  when  they  have  their  homologous 
rides  parallel  or  perpendicular  to  each  other. 

Let  the  triangles  ABC,  abc,  DEF  have  their  homologous 
sides  parallel  or  perpendicular  to  each  other ;  the  triangles 
are  similar. 

First.  Let  the  homologous 
sides  be  parallel  to  each  other. 
If  the  side  AB  is  parallel  to 
ab,  and  BC  to  be,  the  angle  B 
is  equal  to  the  angle  b  (Prop. 
XXVL,  B.  I.) ;  also,  if  AC  is 
parallel  to  ac,  the  angle  C  is 
equal  to  the  angle  c ;  and  hence 

the  angle  A  is  equal  to   the  ^ g- 

angle  a.     Therefore  the  trian- 
gles ABC,  abc  are  equiangular,  and  consequently  similar. 

Secondly.  Let  the  homologous  sides  be  perpendicular  to 
each  other.  Let  the  side  DE  be  perpendicular  to  AB,  and 
the  side  DF  to  AC.  Produce  DE  to  I,  and  DF  to  H ;  then, 
in  the  quadrilateral  AIDH,  the  two  angles  I  and  H  are  right 
angles.  But  the  four  angles  of  a  quadrilateral  are  together 
equal  to  four  right  angles  (Prop.  XXVIIL,  Cor.  1,  B.  I.) ; 
therefore  the  two  remaining  angles  IAH,  IDH  are  together 
equal  to  two  right  angles.  But  the  two  angles  EDF,  IDH 
are  together  equal  to  two  right  angles  (Prop.  II.,  B.  I.); 
therefore  the  angle  EDF  is  equal  to  IAH  or  BAG. 

la  the  same  manner,  if  the  side  EF  is  also  perpendicular  to 
BC,  it  may  be  proved  that  the  angle  DFE  is  equal  to  C,  and, 
consequently,  the  angle  DEF  is  equal  to  B ;  hence  the  trian- 
gles ABC,  DEF  are  equiangular  and  similar.  Therefore,  two 
triangles  &c. 

Scholium.  When  the  sides  of  the  two  triangles  are  para.- 
iel,  the  parallel  sides  are  homologous  ;  but  when  the  sides  are 
perpendicular  to  each  other,  the  perpendicular  sides  are  ho- 
mobgous.  Thus  DE  is  homologous  to  AB,  DF  to  AC,  and 
EF  to  BC 

D 


74  GEOMETRY. 


PROPOSITION   XXII.       THEOREM. 

In  a  right-angled  triangle,  if  a  perpendicular  is  d'  awn  from 
the  right  angle  to  the  hypoihenuzz , 

1st.  The  triangles  on  each  side  of  the  perpendicular  are  sim- 
ilar to  the  whole  triangle  and  to  each  other. 

2d.  The  perpendicular  is  a  mean  proportional  between  th* 
segments  of  the  hypothenuse. 

3d.  Each  of  the  sides  is  a  mean  proportional  between  the  hy 
pothenuse  and  its  segment  adjacent  to  that  side. 

Let  ABC  be  a  right-angled  triangle,  hav-                     A 
ing  the  right  angle  BAG,  and  from  the  angle 
A  let  AD  be  drawn  perpendicular  to  the 
hypothenuse  BC.  ^ . 

First.  The  triangles  ABD,  ACD  are  sim-  B  DO 

ilar  to  the  whole  triangle  ABC,  and  to  each  other. 

The  triangles  BAD,  BAG  have  the  common  angle  B,  also 
the  angle  BAG  equal  to  BDA,  each  of  them  being  a  right  an- 
gle, and,  therefore,  the  remaining  angle  ACB  is  equal  to  the 
remaining  angle  BAD  (Prop.  XXVIL,  Cor.  2,  B.  I.) ;  therefore 
the  triangles  ABC,  ABD  are  equiangular  and  similar.  In 
like  manner,  it  may  be  proved  that  the  triangle  ADC  is  equi 
angular  and  similar  to  the  triangle  ABC  ;  therefore  the  three 
triangles  ABC,  ABD,  ACD  are  equiangular  and  similar  to 
each  other. 

Secojidly.  The  perpendicular  AD  is  a  mean  proportional  be 
tween  the  segments  BD,  DC  of  the  hypothenuse.     For,  sinct 
the  triangle  ABD  is  similar  to  the  triangle  ADC,  their  ho 
mologous  sides  are  proportional  (Def.  3),  and  we  have 
BD  :  AD  :  :  AD  :  DC. 

Thirdly.  Each  of  the  sides  AB,  AC  is  a  mean  proportional 
between  the  hypothenuse  and  the  segment  adjacent  to  that 
side.  For,  since  the  triangle  BAD  is  similar  to  the  triangle 
BAG,  we  have 

BC  :  BA  :  :  BA  :  BD. 

And,  since  the  triangle  ABC  is  similar  to  the  triangle  ACD 
we  have 

BC  :  CA  :  :  CA  :  CD 
Therefore,  in  a  right-angled  triangle,  &c. 

Cor.  If  from  a  point  A,  in  the  circumfe- 
rence  of  a  circle,  two  chords  AB,  AC  are 
drawn  to  the  extremities  of  the  diameter 
BC,  the  triangle  BAG  will  be  right-angled 
at  A  (Prop.  XV.,  Cor.  2,  B.  III.)  ;  therefore  B 


BOOK    IV. 


75 


the  perpendicular  AD  is  a  mean  proportional  between  BD 
and  DC,  the  two  segments  of  the  diameter ;  that  is, 
AD2=BDxDC. 


PROPOSITION    XXIII.       THEOREM. 

Two  triangles,hamng  an  angle  in  the  one  equal  to  an  angle 
tn,  the  other,  are  to  each  other  as  the  rectangles  of  ike  sides 
which  contain  the  equal  angles. 

Let  the  two  triangles  ABC,  ADE  have 
the  angle  A  in  common ;  then  will  the  trian- 
gle ABC  be  to  the  triangle  ADE  as  the  rect- 
angle AB  X  AC  is  to  the  rectangle  AD  X  AE. 

Join  BE.  Then  the  two  triangles  ABE, 
ADE,  having  the  common  vertex  E,  have 
ihe  same  altitude,  and  are  to  each  other  as 
their  bases  AB,  AD  (Prop.  VI.,  Cor.  1)  ; 
therefore 

ABE  :  ADE  :  :  AB  :  AD. 

Also,  the  two  triangles  ABC,  ABE,  having  the  common 
vertex  B,  have  the  same  altitude,  and  are  to  each  other  aa 
their  bases  AC,  AE  ;  therefore 

ABC  :  ABE  :  :  AC  :  AE. 

Hence  (Prop.  XL,  Cor.,  B.  II.). 

ABC  :  ADE  :  :  AB  X  AC  :  AD  X  AE. 

Therefore,  two  triangles,  &c. 

Cor.  1.  If  the  rectangles  of  the  sides  containing  the  equal 
angles  are  equivalent,  the  triangles  will  be  equivalent. 

Cor.  2.  Equiangular  parallelograms  are  to  each  other  as 
trie  rectangles  of  the  sides  which  contain  the  equal  angles. 


PROPOSITION    XXIV.       THEOREM. 

Similar  triangles  are  to  each  other  as  the  squares  described 
on  their  homologous  sides. 

Let  ABC,  DEF  be  two  simi- 
lar triangles,  having  the  angle  A 
equal  to  D,  the  angle  B  equal  to 
E,  and  C  equal  to  F ;  then  the 
triangle  ABC  is  to  the  triangle 
DEF  as  the  square  on  BC  is  to  B 
the  square  on  EF. 

By  similar  triangles,  wre  have  (Def.  3) 
AB  :  DE  :  :  BC  :  EF. 

\lsn.  BC  :  EF  :  :  BC  :  EF. 


76  GEOMETRY 

Multiplying  together  the  corresponding  terms  of  these  pro- 
portions, we  obtain  (Prop.  XL,  B.  II.), 

ABxBC  :  DExEF  :  :  BC2  :  EF2. 
But,  by  Prop.  XXIII., 

ABC  :  DEF  :  :  ABxBC  :  DExEF; 
*ence  (Prop.  IV.,  B.  II.) 

ABC  :  DEF  :  :  BC2  :  EF2. 
Therefore,  similar  triangles,  &c. 


PROPOSITION    XXV.       THEOREM. 

Two  similar  polygons  may  be  divided  into  the  same  numoei 
of  triangles,  similar  each  to  each,  and  similarly  situated. 

Let  ABCDE,  FGHIK 

be  two  similar  polygons  ; 
they  may  be  divided  into 
the  same  number  of  sim- 
ilar triangles.  Join  AC, 
AD,  FH,  FI. 

Because    the    polygon 
ABCDE  is  similar  to  the  B 

polygon  FGHIK,  the  angle  B  is  equal  to  the  angle  G  (Del. 
3),  and  AB  :  BC  :  :  FG  :  GH.  And,  because  the  triangles 
ABC,  FGH  have  an  angle  in  the  one  equcU  to  an  angle  in 
the  other,  and  the  sides  about  these  equal  angles  proportion- 
al, they  are  similar  (Prop.  XX.) ;  therefore  the  angle  BCA 
is  equal  to  the  angle  GHF.  Also,  because  the  polygons  are 
similar,  the  whole  angle  BCD  is  equal  (Def.  3)  to  the  whole 
angle  GHI ;  therefore,  the  remaining  angle  ACD  i-s  equal  to 
the  remaining  angle  FHI.  Now,  because  the  triangles  ABC 
FGH  are  similar, 

AC  :  FH  :  :  BC  :  GH. 
And,  because  the  polygons  are  similar  (Def.  3), 

BC  :  GH  :  :  CD  :  HI ; 
whence  AC  :  FH  :  :  CD  :  HI ; 

that  is,  the  sides  about  the  equal  angles  ACD,  FHI  are  pro* 
portional ;  therefore  the  triangle  ACD  is  similar  to  the  trian- 
gle FHI  (Prop.  XX.).  For  the  same  reason,  the  triangle 
ADE  is  similar  to  the  triangle  FIK;  therefore  the  similar 
polygons  ABCDE,  FGHIK  are  divided  into  the  same  num- 
ber of  triangles,  which  are  similar,  each  to  each,  and  similar- 
ly situated. 

Cor.  Conversely,  if  two  polygons  are  composed  of  the  same 
number  of  triangles,  similar  and  similarly  situated  the  poly- 
gons we  similar 


B03R    IV.  71 

For,  because  the  triangles  are  similar,  the  angle  ABC  is 
equal  to  FGH  ;  and  because  the  angle  BCA  is  equal  to  GHF 
and  ACD  to  FHI,  therefore  the  angle  BCD  is  equal  to  GHI 
For  the  same  reason,  the  angle  CDE  is  equal  to  HIK,  and  so 
on  for  the  other  angles.  Therefore  the  two  polygons  are  mu- 
tually equiangular. 

Moreover,  the  sides  about  the  equal  angles  are  proportion- 
al. For,  because  the  triangles  are  similar,  AB  :  FG  :  :  BC  : 
GH.  Also,BC  :  GH  :  :  AC  :  FH,and  AC  :  FH  :  :  CD  :  HI; 
hence  BC  :  GH  :  :  CD  :  HI.  In  the  same  manner,  it  may  be 
proved  that  CD  :  HI  :  :  DE  :  IK,  and  so  on  for  the  other 
sides.  Therefore  the  two  polygons  are  similar. 


PROPOSITION  XXVI.       THEOREM. 

The  perimeters  of  similar  polygons  are  to  each  other  as  then 
homologous  sides;  and  their  areas  are  as  the  squares  of  those 
sides. 

Let  ABCDE,  FGHIK 

be  two  similar  polygons, 

and  let  AB  be  the  side 

homologous' to  FG;  then 

the  perimeter  of  ABCDE 

is    to    the    perimeter    of 

FGHIK  as  AB  is  to  FG ; 

and  the  area  of  ABCDE  E 

is  to  the  area  of  FGHIK  as  AB2  is  to  FG2 

First.  Because  the  polygon  ABCDE  is  similar  to  the  pol- 
ygon FGHIK  (Def.  3), 

AB  :  FG :  :  BC  :  GH  :  :  CD  :  HI,  &c.; 
therefore  (Prop.  IX.,  B.  II.)  the  sum  of  the  antecedents  AB 
f  BC-f  CD,  &c.,  which  form  the  perimeter  of  the  first  figure 
is  to  the  sum  of  the  consequents  FG-fGH+HI,  &c.,  which 
form  the  perimeter  of  the  second  figure,  as  any  one  antece- 
dent is  to  its  consequent,  or  as  AB  to  FG. 

Secondly,  Because  the  triangle  ABC  is  similar  to  the-  tri. 
angle  FGH,  the  triangle  ABC":  triangle  FGH  :  :  AC2  :  FHa 
(Prop.  XXIV.). 

And,  because  the  triangle  ACD  is  similar  to  the  triangle 
FHI, 

ACD  :  FHI  :  :  AC2  :  FH2. 

Therefore  the  triangle  ABC  :  triangle  FGH  :  :  triangle 
ACD  :  triangle  FHI  (Prop.  IV.,  B.  II.).  In  the  same  man- 
ner,  it  may  be  proved  that 

ACD  :  FHI  :  :  ADE  :  FIK. 


GEOMETRY. 


Therefore,   as    Jie   sum   of  the   antecedents   ABC+ACD-f 
ADE,  or  the  polygon  ABODE,  is  to  the  sum  of  the  conse- 
quents FGH+FHI+FIK,  or  the  polygon  FGHIK,  so  is  any 
one  antecedent,  as  ABC,  to  its  consequent  FGH ;  or,  as  AB* 
to  FG2.     Therefore,  similar  polygons,  &o. 


c 


PROPOSITION    XXVII.       THEOREM. 

If  two  chords  in  a  circle  intersect  each  other,  the  rectangle 
contained  by  the  parts  of  the  one,  is  equal  to  the  rectangle  con- 
tained by  the  parts  of  the  other. 

Let  the  two  chords  AB,  CD  in  the  circle 
ACBD,  intersect  each  other  in  the  point  E ; 
the  rectangle  contained  by  AE,  EB  is  equal 
to  the  rectangle  contained  by  DE,  EC. 

Join  AC  and  BD.  Then,  in  the  triangles 
ACE,  DBE,  the  angles  at  E  are  equal,  be- 
ing vertical  angles  (Prop.  V.,  B.  I.)  ;  the 
angle  A  is  equal  to  the  angle  D,  being  in- 
scribed in  the  same  segment  (Prop.  XV.,  Cor.  1.,  B.  III.) ; 
therefore  the  angle  C  is  equal  to  the  angle  B.  The  triangle* 
are  consequently  similar ;  and  hence  (Prop.  XVIII.) 

AE  :  DE  :  :  EC  :  EB, 
or  (Prop.  I.,  B.  II.), 

AExEB-DExEC. 
Therefore,  if  two  chords,  &c. 

Cor.  The  parts  of  two  chords  which  intersect  each  other  in 
a  circle  are  reciprocally  proportional;  that  is,  AE  :  DE  :  : 
EC  :  EB. 


PROPOSITION    XXVIII.       THEOREM. 

If  from  a  point  without  a  circle,  a  tangent  and  a  secant  be 
drawn,  the  square  of  the  tangent  will  be  equivalent  to  the  red 
angle  contained  by  the  whole  secant  and  its  external  segment. 

Let  A  be  any  point  without  the  circle 
BCD,  and  let  AB  be  a  tangent,  and  AC  a 
secant ;  then  the  square  of  AB  is  equiva- 
lent to  the  rectangle  AD  X  AC. 

Join  BD  and  BC.  Then  the  triangles 
ABD  and  ABC  are  similar ;  because  they 
have  the  angle  A  in  common ;  also,  the 
angle  ABD  formed  by  a  tangent  and  a 
chord  is  measured  by  half  the  arc  BD 


BOOK    IV.  79 

(Prop,  XVI.,  B.  III.) ;  and  the  angle  C  is  measured  by  half 
the  same  arc,  therefore  the  angle  ABD  is  equal  to  C,  and  the 
two  triangles  ABD,  ABC  are  equiangular,  and,  consequently 
similar  •  therefore  (Prop.  XVIII.) 

AC  :  AB  :  :  AB  :  AD ; 
whence  (Prop.  I.,  B.  II.), 

AB2=ACxAD. 
Therefore,  if  from  a  point,  &c. 

Cor.  1.  If  from  a  point  without  a  circle,  a  tangent  and  a  se- 
cant be  drawn,  the  tangent  will  be  a  mean  proportional  be- 
tween the  secant  and  its  external  segment. 

Cor.  2.  If  from  a  point  without  a  circle,  two  secants  be 
drawn,  the  rectangles  contained  by  the  whole  secants  and 
their  external  segments  will  be  equivalent  to  each  other ;  foi 
each  of  these  rectangles  is  equivalent  to  the  square  of  the 
tangent  from  the  same  point. 

Cor.  3.  If  from  a  point  without  a  circle,  two  secants  be 
drawn,  the  whole  secants  will  be  reciprocally  proportional  to 
their  external  segments. 

PROPOSITION  XXIX.       THEOREM. 

If  an  angle  of  a  triangle  be  bisected  by  a  line  which  cuts  t/ie 
base,  the  rectangle  contained  by  the  sides  of  the  triangle,  is 
equivalent  to  the  rectangle  contained  by  the  segments  of  thp 
base,  together  with  the  square  of  the  bisecting  line. 

Let  ABC  be  a  triangle,  and  let  the  an- 
gle BAG  be  bisected  by  the  straight  line 
AD  ;  the  rectangle  B  A  X  AC  is  equiva- 
lent to  BD  X  DC  together  with  the  square 
of  AD. 

Describe  the  circle  ACEB  about  the 
triangle,  and  produce  AD  to  meet  the  cir- 
cumference in  E,  and  join  EC.  Then,  be- 
cause the  angle  BAD  is  equal  to  the  an- 
gle CAE,  and  the  angle  ABD  to  the  angle  AEC,  for  they  are 
in  the  same  segment  (Prop.  XV.,  Cor.  1,  B.  III.),  the  trian- 
gles ABD,  AEC  are  mutually  equiangular  and  similar ;  there- 
fore (Prop.  XVIII.) 

BA  :  AD  :  :  EA  :  AC ; 
consequently  (Prop.  I.,  B.  II.), 

BAxAC=ADxAE. 

But  AE=AD-f-DE  ;  and  multiplying  each  of  these  equals 
by  AD,  we  have    (Prop.  III.)   ADx  AE  =  AD'H-ADxDE. 
But  ADxDE=BDxDC  (Prop.  XXVII.);  hence 
BAxAC=BDxDC-J-AD'. 

Therefore,  if  an  angle,  &c 


80  GEOMETRY. 


PROPOSITION   XXX.       THEOREM. 

The  rectangle  contained  by  the  diagonals  of  a  quadrilateral 
inscribed  in  a  circle,  is  equivalent  to  the  sum  of  the  rectangle* 
of  the  opposite  sides. 

Let  ABCD  be  any  quadrilateral  in- 
scribed in  a  circle,  and  let  the  diagonals 
AC,  BD  be  drawn;  the  rectangle  ACx 
BD  is  equivalent  to  the  sum  of  the  two 
rectangles  ADxBC  and  ABxCD. 

Draw  the  straight  line  BE,  making  the 
angle  ABE  equal  to  the  angle  DBG.  To 
each  of  these  equals  add  the  angle  EBD  ; 
then  will  the  angle  ABD  be  equal  to  the  angle  EBC.  But 
the  angle  BDA  is  equal  to  the  angle  BCE,  because  they  are 
both  in  the  same  segment  (Prop.  XV.,  Cor.  1,  B.  III.) ;  hence 
the  triangle  ABD  is  equiangular  and  similar  to  the  triangle 
EBC.  Therefore  we  have 

AD  :  BD  :  :  CE  :  BC ; 
and,  consequently,  ADxBC=BDxCE. 

Again,  because  the  angle  ABE  is  equal  to  the  angle  DBC 
and  the  angle  BAE  to  the  angle  BDC,  being  angles  in  the 
same  segment,  the  triangle  ABE  is  similar  to  the  triangle 
DBC ;  and  hence 

AB  :  AE  :  :  BD  :  CD ; 
consequently,          AB  X  CD  =BD  X  AE. 
Adding  together  these  two  results,  we  obtain 

ADxBC+ABxCD=BDxCE+BDxAE, 
which  equals      BD  x  (CE+AE),  or  BD  X  AC. 

Therefore,  the  rectangle,  &c. 


PROPOSITION    XXXI.       THEOREM. 

If  from  any  angle  of  a  triangle,  a  perpendicular  be  drawn 
to  the  opposite  side  or  base,  the  rectangle  contained  by  the  sum 
and  difference  of  the  other  two  sides,  is  equivalent  to  the  rect- 
angle contained  by  the  sum  and  difference  of  the  segments  of 
the  base 

Let  ABC  be  any  triangle,  and  let  AD  be  a  perpendicular 
drawn  from  the  angle  A  on  the  base  BC ;  then 

(AC+AB)  x  (AC-AB)  =  (CD+DB)  x  (CD-DB). 
From  A  as  a  center,  with  a  radius  equal  to  AB,  the  short- 


BOOK    IV. 


81 


cr  of  the  two  sides,  describe  a  circumference  BFE.  Pro- 
duce AC  to  meet  the  circumference  in  E,  and  CB,  if  neces- 
sary, to  meet  it  in  F. 

Then,  because  AB  is  equal  to  AE  or  AG,  CE=AC+AB, 
the  sum  of  the  sides  ;  and  CG=AC — AB,  the  difference  of  the 
sides.  Also,  because  BD  is  equal  to  DF  (Prop.  VI.,  B.  III.) ; 
when  the  perpendicular  falls  within  the  triangle,  CF= CD — 
DF=CD— DB,  the  difference  of  the  segments  of  the  base. 
But  when  the  perpendicular  falls  without  the  triangle,  CF= 
CD+DF=CD+DB,  the  sum  of  the  segments  of  the  base. 

Now  in  either  case,  the  rectangle  CE  X  CG  is  equivalent 
to  CBxCF  (Prop.  XXVIIL,  Cor.  2) ;  that  is, 

(AC  +  AB)  x  (AC  -  AB)  =  (CD  +  DB)  x  (CD  -DB). 

Therefore,  if  from  any  angle,  &c. 

Cor.  If  we  reduce  the  preceding  equation  to  a  proportion 
(Prop.  II.,  B.  II.),  we  shall  have 

BC  :  AC  +  AB  :  :  AC-AB  :  CD-DB; 
that  is,  the  base  of  any  triangle  is  to  the  sum  of  the  two  other 
sides,  as  the  difference  of  the  latter  is  to  the  difference  of  the 
segments  of  the  base  made  by  the  perpendicular. 


PROPOSITION    XXXII.       THEOREM. 

The  diagonal  and  side  of  a  square  have  no  common  measure. 

Let  ABCD  be  a  square,  and  AC  its 
diagonal ;  AC  and  AB  have  no  common 
measure. 

In  order  to  find  the  common  measure, 
if  there  is  one,  we  must  apply  CB  to  CA 
as  often  as  it  is  contained  in  it.  For  this 
purpose,  from  the  center  C,  with  a  radius 
CB,  describe  the  semicircle  EBF.  We  A  G 
perceive  that  CB  is  contained  once  in  AC,  with  a  remainder 
AE,  which  remainder  must  be  compared  wi.h  BC  or  its 
equal  AB. 

Now,  since  the  angle  ABC  is  a  right  angle,  AB  is  a  tan- 
gent to    he  circumference  ;  and  AE  :  AB  :  :  AB  :  AF  ('*  rop. 


GEOMETRY. 


XXV1IL,  Oor.  1).  Instead,  therefore,  of 
comparing  AE  with  AB,  we  may  substi- 
tute the  equal  ratio  of  AB  to  AF.  But 
AB  is  contained  twice  in  AF,  with  a  re- 
mainder AE,  which  must  be  again  com- 
pared with  AB.  Instead,  however,  of 
comparing  AE  with  AB,  we  may  again 
employ  the  equal  ratio  of  AB  to  AF. 
Hence  at  each  operation  we  are  obliged  to  compare  AB  with 
AF,  which  leaves  a  remainder  AE  ;  from  which  we  see  that 
the  process  will  never  terminate,  and  therefore  there  is  no 
common  measure  between  the  diagonal  and  side  of  a  square 
that  is,  there  's  no  line  which  is  contained  an  exact  number 
of  times  in  each  of  them. 


BOOK  v  88 


BOOK  V 

PROBLEMS 
Postulates. 

\    A  /".raight  line  may  be  drawn  from  any  one  point  to 
any  other  point.    . 

2.  A  terminated  straight  line  may  be  produced  to  any 
length  in  a  straight  line. 

3.  From  the  greater  of  two  straight  lines,  a  part  may  be 
cut  off  equal  to  the  less. 

4.  A  circumference  may  be  described  from  any  center,  and 
with  any  radius. 


PROBLEM   I. 

To  bisect  a  given  straight  line. 

Let   AB   be   the   given   straight   line 
which  it  is  required  to  bisect.    • 

From  the  center  A,  with  a  radius  great- 
er than  the  half  of  AB,  describe  an  arc  of 
a  circle  (Postulate  4)  ;  and  from  the  cen- 
ter B,  with  the  same  radius,  describe  an- 
other arc  intersecting  the  former  in  D  and 
E.  Through  the  points  of  intersection,  draw  the  straight  line 
DE  (Post.  1)  ;  it  will  bisect  AB  in  C. 

For,  the  two  points  D  and  E,  being  each  equally  distanl 
from  the  extremities  A  and  B,  must  both  lie  in  the  perpen- 
dicular, raised  from  the  middle  point  of  AB  (Prop.  XVIII. 
Cor.,  B.  I.).  Therefore  the  line  DE  divides  the  line  AB 
into  two  equai  parts  at  the  point  C. 


84  GEOMETRY. 


PROBLEM   II. 

To  draw  a  perpendiculai  to  a  straight  line,  from  a  given 
point  in  tint  line. 

Let  BC  be  the  given  straight  line,  and  A 
the  point  given  in  it ;  it  is  required  to  draw 
a  straight  line  perpendicular  to  BC  through          .  ' 
the  given  poin.  A. 

In  the  straight  line  BC  take  any  point  B 

and  make  AC  equal  to  AB  (Post.  3).     From  ^ 

B  as  a  center,  with  a  radius  greater  than 
BA,  describe  an  arc  of  a  circle  (Post.  4)  ;  and  from  C  as  a 
center,  with  the  same  radius,  describe  another  arc  intersect- 
ing the  former  in  D.  Draw  AD  (Post.  1),  and  it  will  be  the 
perpendicular  required. 

For,  the  points  A  and  D,  being  equally  distant  from  B  and 
C,  must  be  in  a  line  perpendicular  to  the  middle  of  BC  (Prop. 
XVIIL,  Cor.,  R  L).  Therefore  AD  has  been  drawn  per- 
pendicular to  BC  from  the  point  A. 

Scholium.  The  same  construction  serves  to  make  a  right 
angle  BAD  at  a  given  point  A,  on  a  given  line  BC. 


PROBLEM    III. 

To  draw  a  perpendicular-  to  a  straight  line,  from  a  given 
point  without  it. 

Let  BD  bo  a  straight  line  of  unlimited  A 

length,  and  let  A  be  a  given  point  without 
it.  It  is  required  to  draw  a  perpendicular 
to  BD  from  the  point  A. 

Take  any  point  E  upon  the  other  side 


of  BD ;  and  from  the  center  A,  with  the        ^- -- — -^ D 
radius  AE,  describe  the  arc  BD  cutting 
the  line  BCD  in  the  two  points  B  and  D. 
From  the  points  B  and  D  as  centers,  de- 
scribe two  arcs,  as  in  Prob.  II.,  cutting  each  other  in  F. 
Join  AF,  and  it  will  be  the  perpendicular  required. 

For  the  two  points  A  and  F  are  each  equally  distant  from 
the  points  B  and  D ;  therefore  the  line  AF  has  been  drawn 
perpendicular  to  BD  (Prop.  XVIIL,  Cor.,  B.  L),  from  the 
given  point  A. 


BOOK    V 


B  C 


PROBLEM   IV. 

At  a  given  point  in  a  straight  line,  if  make  an  angle  cquctt 
U  a  given  angle. 

Let  AB  be  the  given  straight 
line.  A  the  given  point  in  it,  and 
C  the  given  angle  ;  it  is  required 
to  make  an  angle  at  the  point  A 
in  the  straight  line  AB,  that  shall  A 
be  equal  to  the  given  angle  C. 

With  C  as  a  center,  and  any  radius,  describe  an  arc  DE 
terminating  in  the  sides  of  the  angle ;  and  from  the  point  A 
as  a  center,  with  the  same  radius,  describe  the  indefinite  arc 
BF.  Draw  the  chord  DE ;  and  from  B  as  a  center,  with  a 
radius  equal  to  DE,  describe  an  arc  cutting  the  arc  BF  in  G. 
Draw  AG,  and  the  angle  BAG  will  be  equal  to  the  given 
angle  C. 

For  the  two  arcs  BG,  DE  are  described  with  equal  radii, 
and  they  have  equal  chords  ;  they  are,  therefore,  equal  (Prop. 
III.,  B.  III.).  But  equal  arcs  subtend  equal  angles  (Prop 
IV.,  B.  III.)  ;  and  hence  the  angle  A  has  been  made  equal  to 
the  given  angle  C. 


PROBLEM    V. 

To  bisect  a  given  arc  or  angle. 

First.  Let  ADB  be  the  given  arc  which 
it  is  required  to  bisect. 

Draw  the  chord  AB,  and  from  the  center 
C  draw  CD  perpendicular  to  AB  (Prob. 
III.)  ;  it  will  bisect  the  arc  ADB  (Prop. 
VI.,  B.  III.),  because  CD  is  a  radius  per- 
pendicular to  a  chord. 

Secondly.  Let  ACB  be  an  angle  which  it  is  required  to  bi- 
sect. From  C  as  a  center,  with  any  radius,  describe  an  arc 
AB ;  and,  by  the  first  case,  draw  the  line  CD  bisecting  the 
arc  ADB.  The  line  CD  will  also  bisect  the  angle  ACB.  For 
the  angles  ACD,  BCD  are  equal,  being  subtended  by  the 
equal  arcs  AD,  DB  (Prop.  IV.,  B.  III.). 

Scholium.  By  the  same  construction,  each  of  the  halves 
AD,  DB  may  be  bisected ;  and  thus  by  successive  bisections 
an  arc  or  angle  may  be  divide  1  into  four  equal  parts,  intc 
eight,  sixteen,  &c. 


86  GEOMETRY. 


PROBLEM  VI. 

Thrcugh  %  given  point,  to  draw  a  straight  line  parallel  to 
a  given  line. 

Let  A  be  the  gi  /en  point,  and  BC  the  -g                  DC 
given  straight  line ;  it  is  required  to  draw                  ~s- 
through  the  goint  A,  a  straight  line  paral- 
lel to  BC. ,  -^ g 

In  BC  take  any  point  D,  and  join  AD. 

Then  at  the  point  A,  in  the  straight  line  AD,  make  the  angle 
DAE  equal  to  the  angle  ADB  (Prob.  IV.). 

Now,  because  the  straight  line  AD,  which  meets  the  ,twc 
straight  lines  BC,  AE,  makes  the  alternate  angles  ADB,  DAE 
equal  to  each  other,  AE  is  parallel  to  BC  (Prop.  XXII. ,  B. 
I.).  Therefore  the  straight  line  AE  has  been  drawn  through 
the  point  A,  parallel  to  the  given  line  BC. 


PROBLEM  VII. 

Two  angles  of  a  triangle  being  given,  to  find  the  third  angle. 

The  three  angles  of  every  triangle  are  to- 
gether equal  to  two  right  angles  (Prop. 
XXVIL,  B.  I.).  Therefore,  draw  the  in- 
definite line  ABC.  At  the  point  B  make 
the  angle  ABD  equal  to  one  of  the  given  ADC 
angles  (Prob.  IV.),  and  the  angle  DBE  equal  to  the  other 
given  angle ;  then  will  the  angle  EBC  be  equal  to  the  third 
angle  of  the  triangle.  For  the  three  angles  ABD.  DBE, 
EBC  are  together  equal  to  two  right  angles  (Prop.  II.,  B 
I.),  which  is  the  sum  of  all  the  angles  of  the  triangle. 

PROBLEM  VIII. 

Given  two  sides  and  the  included  angle  of  a  triangle,  to  con 
struct  the  triangle. 

Draw  the  straight  line  BC  equal  to  one  A 

of  the  given  sides.  At  the  point  B  make 
the  angle  ABC  equal  to  the  given  angle 
(Prob.  IV.) ;  and  take  AB  equal  to  the  other 
side.  Join  AC.  and  ARC  will  he.  the 


BOOK    V  87 


triangle  required.  For  its  sides  AB,  BC  are  made  equal  tc 
the  given  sides,  and  the  included  angle  B  is  made  equal  to 
the  given  angle. 


PROBLEM    IX. 

Given  one  side  and  two  angles  of  a  triangle,  to  construct  tfo 
triangle. 

The  two  given  angles  will  either  be  both  adjacent  to  the 
given  side,  or  one  adjacent  and  the  other  opposite.  In  the 
latter  case,  find  the  third  angle  (Prob.  VII.)  ;  and  then  the 
two  adjacent  angles  will  be. known. 

Draw  the  straight  line  AB  equal  to  the 
given  side ;  at  the  point  A  make  the  angle 
BAG  equal  to  one  of  the  adjacent  angles ; 
and  at  the  point  B  make  the  angle  ABD 
equal  to  the  other  adjacent  angle.  The  tw6 
lines  AC,  BD  will  cut  each  other  in  E,  and  J 
ABE  will  be  the  triangle  required ;  for  its  side  AB  is  equal 
to  the  given  side,  and  two  of  its  angles  are  equal  to  the  given 
angles. 


PROBLEM    X. 

Given  the  three  sides  of  a  triangle,  to  construct  the  triangle 

Draw  the  straight  line  BC  equal  to  one  of 
the  given  sides.     From  the  point  B  as  a  cen- 
ter, with  a  radius  equal  to  one  of  the  other 
sides,  describe  an  arc  of  a  circle ;  and  from 
the  point  C  as  a  center,  with  a  radius  equal 
to  the  third  side,  describe  another  arc  cutting      ^_____ 
the  former  in  A.     Draw  AB,  AC ;  then  will  3 
ABC  be  the  triangle  required,  because  its  three  sides  are 
equal  to  the  three  given  straight  lines. 

Scholium.  If  one  of  the  given  lines  was  greater  than  the 
sum  of  the  other  two,  the  arcs  would  not  intersect  each  other, 
and  the  problem  would  be  impossible ;  but  the  solution  will 
always  be  possible  when  the  sum  of  any  two  sides  is  gi  eater 
than  the  third. 


88  GEOMETRY. 


PROBLEM    XT. 

Given  two  sides  of  a  triangle,  and  an  angle  opposite  on*  cj 
tnem,  to  construct  the  triangle. 

Draw  an  indefinite  straight  line 
BC.  At  the  point  B  make  the  angle 
ABC  equal  to  the  given  angle,  and 
make  BA  «qual  to  that  side  which  is 
adjacent  to  the  given  angle.  Then 
from  A  as  a  center,  with  a  radius 
equal  to  the  other  side,  describe  an  arc  cutting  BC  in  the 
points  E  and  F.  Join  AE,  AF.  If  the  points  E  and  F  both 
fall  on  the  same  side  of  the  angle  B,  each  of  the  triangles 
ABE,  ABF  will  satisfy  the  given  conditions  ;  but  if  they  fall 
upon  different  sides  of  B,  only  one  of  them,  as  ABF,  will 
satisfy  the  conditions,  and  therefore  this  will  be  the  triangle 
required. 

If  the  points  E  and  F  coincide  with  one  another,  which 
will  happen  when  AEB  is  a  right  angle,  there  will  be  only 
one  triangle  ABD,  which  is  the  triangle  required. 

Scholium.  If  the  side  opposite  the  given  angle  were  less 
than  the  perpendicular  let  fall  from  A  upon  BC,  the  problem 
would  be  impossible. 


PROBLEM    XII. 

Given  two  adjacent  sides  of  a  parallelogram,  and  the  in- 
cluded angle,  to  construct  the  parallelogram. 

Draw  the  straight  line  AB  equal  to 
one  of  the  given  sides.  At  the  point  A 
make  the  angle  BAG  equal  to  the 
given  angle ;  and  take  AC  equal  to 
the  other  given  side.  From  the  point 
(/  as  a  center,  with  a  radius  equal  to  A 
AB,  describe  an  arc ;  and  from  the  point  B  as  a  center,  with 
a  radius  equal  to  AC,  describe  another  arc  intersecting  the 
former  in  D.  Draw  BD,  CD  ;  then  will  ABDC  be  the  paral- 
lelogram required. 

For,  by  construction,  the  opposite  sides  are  equal ;  there- 
rore  the  figure  is  a  parallelogram  (Prop.  XXX.,  B.  I.),  and  it 
s  formed  with  the  given  sides  and  the  given  angle 


BOOK    V 


Cor.  If  the  given  angle  is  a  rig  at  ait/gle,  the  figure  will  be 
a  rectangle ;  and  if,  at  the  same  time,  the  sides  are  equal,  it 
will  be  a  square. 


PROBLEM    XIII. 

To  find  the  center  of  a  given  circle  or  arc. 

Let  ABC  be  the  given  circle  or  arc ; 
it  is  required  to  find  'ts  center. 

Take  any  three  points  in  the  are,  as 
A  B,  C,  and  join  AB,  BC.  Bisect  AB 
in  1)  (Prob.  L),  and  through  D  draw  DF  A' 
perpendicular  to  AB  (Prob.  II.).  In  the 
same  manner,  draw  EF  perpendicular  to 
BC  at  its  middle  point.  The  perpen- 
diculars DF,  EF  will  meet  in  a  point  F  equally  distant  from 
the  points  A,  B,  and  C  (Prop.  VII.,  B.  III.)  ;  and  therefore  F  ia 
the  center  of  the  circle. 

Scholium.  By  the  same  construction,  a  circumference  may 
be  made  to  pass  through  three  given  points  A,  B,  C ;  and 
also,  3.  circle  may  be  described  about  a  triangle. 


PROBLEM    XIV. 


Through  a  given  point,  to  draw  a  tangent  to  a  given  circle 

First.  Let  the  given  point  A  be 
without  the  circle  BDE  ;  it  is  re- 
quired to  draw  a  tangent  to  the  cir- 
cle through  the  point  A. 

Find  the  center  of  the  circle  C,  and 
join  AC.     Bisect  AC  in  D  ;  and  with 
D  as  a  center,  and  a  radius  equal  to 
AD,  describe  a  circumference  intersecting  the  given  circuin 
ference  in  B.     Draw  AB,  and  it  will  be  the  tangent  required. 

Draw  the  radius  CB.  The  angle  ABC,  being  inscribed  in 
a  semicircle  is  a  right  angle  (Prop.  XV.,  Cor.  2,  B.  III.). 
Hence  the  line  AB  is  a  perpendicular  at  the  extremity  of  the 
radius  CB ;  it  is,  therefore,  a  tangent  to  the  circumference 
(Prop  IX.,  B.  III.). 

Secondly.  If  the  given  point  is  in  the  circumference  of  the 
.  circle,  as  the  point  B,  draw  the  radius  BC,  and  make  BA 
perpendicular  to  BC,      BA   will   be   the   tangent   required 
(Prop.  IX.,  B.  III.). 


90  GEOMETRY. 

Schdium.  When  the  point  A  lies  without  the  circle,  two 
tangents  may  always  be  drawn;  for  the  circumference  whose 
center  is  D  intersects  the  given  circumference  in  two  points. 


PROBLEM    XV. 

To  inscribe  a  circle  in  a  given  triangle. 

Let  ABC  be  the  given  triangle  ;  it  is 
required  to  inscribe  a  circle  in  it. 

Bisect  the  angles  B  and  C  by  the 
ines  BD,  CD,  meeting  each  other  in 
the  point  D.  From  the  point  of  inter- 
section, let  fall  the  perpendiculars  DE, 
DF,  DG  on  the  three  sides  of  the  tri- 
angle ;  these  perpendiculars  will  all  be 
equal.  For,  by  construction,  the  angle  B 
EBD  is  equal  to  the  angle  FED ;  the  right  angle  DEB  is 
equal  to  the  right  angle  DFB ;  hence  the  third  angle  BDE 
is  equal  to  the  third  angle  BDF  (Prop.  XXVIL,  Cor.  2,  B. 
L).  Moreover,  the  side  BD  is  common  to  the  two  triangles 
BDE,  BDF,  and  the  angles  adjacent  to  the  common  side  are 
equal ;  therefore  the  two  triangles  are  equal,  and  DE  is  equal 
to  DF.  For  the  same  reason,  DG  is  equal  to  DF.  There- 
fore the  three  straight  lines  DE,  DF,  DG  are  equal  to  each 
other ;  and  if  a  circumference  be  described  from  the  center 
D,  with  a  radius  equal  to  DE,  it  will  pass  through  the  ex- 
tremities of  the  lines  DF,  DG.  It  will  also  touch  the  straight 
lines  AB,  BC,  CA,  because  the  angles  at  the  points  E,  F,  G 
are  right  angles  (Prop.  IX.,  B.  III.).  Therefore  the  circle 
EFG  is  inscribed  in  the  triangle  ABC  (Def.  11,  B.  III.) 

Scholium.     The  three  lines  which  bisect  the  angles  of  a 
triangle,  all  meet  in  the  same  point,  viz.,  the  center  of  the  in 
scribed  circle. 


PROBLEM    XVI. 

Upon  a  given  straight  line,  to  describe  a  segment  of  a  circle, 
which  shall  contain  a  given  angle. 

Let  AB  be  the  given  straight  line,  upon  which  it  is  re- 
quired to  describe  a  segment  of  a  circle  containing  a  given 
angle. 

At  the  point  A,  in  the  straight  line  AB,  make  the  angle 
SAD  equal  to  the  given  angle ;  and  from  the  point  A  draw 


BOOK    V 


91 


D'  F 

AC  psrpendicular  to  AD.     Bisect  AB  in  E,  and  from  E 
draw  EC  perpendicular  to  AB.     From  the  point  C,  where 
these  perpendiculars  meet,  with  a  radius  equal  to  AC,  de 
scribe  a  circle.     Then  will  AGB  be  the  segment  required. 

For,  since  AD  is  a  perpendicular  at  the^  extremity  of  the 
radius  AC,  it  is  a  tangent  (Prop.  IX.,  B.  III.) ;  and  the  angle 
BAD  is  measured  by  half  the  arc  AFB  (Prop.  XVI.,  B.  III.). 
Also,  the  angle  AGB,  being  an  inscribed  angle,  is  measured 
by  half  the  same  arc  AFB ;  hence  the  angle  AGB  is  equal  to 
the  angle  BAD,  which,  by  construction,  is  equal  to  the  given 
angle.  Therefore  all  the  angles  inscribed  in  the  segment 
A.GB  are  equal  to  the  given  angle. 

Scholium.  If  the  given  angle  was  a  right  angle,  the  re- 
quired segment  would  be  a  semicircle,  described  on  AB  as  a 
diameter. 


PROBLEM    XVII. 

To  divide  a  given  straight  line  into  any  number  of  equal 
parts,  or  into  parts  proportional  to  given  lines. 

First.  Let  AB  be  the  given  straight 
line  which  it  is  proposed  to  divide  into 
any  number  of  equal  parts,  as,  for  ex- 
ample, five. 

From  the  point  A  draw  the  indefinite 
straight  line  AC,  making   any  angle 
with  AB.      In  AC  take  any  point  D, 
and  set  off  AD  five  times  upon  AC.     Join  BC,  and  draw  DE 
parallel  to  it ;  then  is  AE  the  fifth  part  of  AB. 

For,  since  ED  is  parallel  to  BC,  AE  :  AB  :  :  AD  :  AC 
(Prop.  XVI.,  B.  IV.).  But  AD  is  the  fifth  part  of  AC ; 
therefore  AE  is  the  fifth  part  of  AB. 

Secondly.  Let  AB  be  the  given  straight  line,  and  AC  a  di- 
vided line ;  it  is  required  to  divide  AB  similarly  to  AC.    Sup- 
'  pose  AC  *o  be  divided  in  the  points  D  and  E.     Place  AB, 
AC  so  as  to  contain  any  angle ;  join  BC,  and  through  the 


92 


GEOMETRl 


points  D,  E  draw  DF,  EG  parallel  to  BC. 
The  line  AB  wil.  be  divided  into  parts 
proportional  to  those  of  AC. 

For,  because  DF  and  EG  are  both  par- 
allel to  CB,  we  have  AD  :  AF  : :  DE  :  FG 
:  EC :  GB  (Prop.  XVI.,  Cor.  2,  B.  IV.).      A 


G     l] 


PROBLEM    XVIII. 


To  find  a  fourth  proportional  to  three  given  lines. 

From  any  point  A  draw  two  straight 
lines  AD,  AE,  containing  any  angle 
DAE ;  and  make  AB,  BD,  AC  respect- 
ively equal  to  the  proposed  lines.  Join 
B,  C ;  and  through  D  draw  DE  parallel 
to  BC ;  then  will  CE  be  the  fourth  pro- 
portional required. 

For,  because  BC  is  parallel  to  DE,  we  have 

AB  :  BD  :  :  AC  :  CE  (Prop.  XVL,  B.  IV.). 

Cor.  In  the  same  manner  may  be  found  a  third  propor 
tional  to  two  given  lines  A  and  B ;  for  this  will  be  the  same 
as  a  fourth  proportional  to  the  three  lines  A,  B,  B. 


PROBLEM    XIX. 


To  find  a  mean  proportional  between  two  given  lines. 


.,.. I> 


/ 


Let  AB,  BC  be  the  two  given  straight 
lines ;  it  is  required  to  find  a  mean  pro- 
portional between  them. 

Place  AB,  BC  in  a  straight  line ;  upon 
AC  describe  the  semicircle  ADC ;   and 
from  the  point  B  draw  BD  perpendicular  J^ 
to  AC.     Then  will  BD  be  the  mean  proportional  required. 

For  the  perpendicular  BD,  let  fall  from  a  point  in  the  cir- 
cumference upon  the  diameter,  is  a  mean  proportional  be- 
tween the  two  segments  of  the  diameter  AB,  BC  (Prop. 
XXII.,  Cor.,  B.  IV.) ;    and  these  segments  are  equal  to  tha 
wo  given  lines, 


BOOK    Y.  93 


PROBLEM    XX. 

To  divide  a  given  line  into  two  parts,  such  that  ihe  greater 
part  may  be  a  mean  proportional  between  the  whole  line  and 
the  other  part. 

Let  AB  be  the  given  straight  line ; 
it  is  required  to  divide  it  into  two 
parts  at  the  point  F,  such  that  AB  : 
AF  :  :  AF  :  FB. 

At  the  extremity  of  the  line  AB, 
erect  the  perpendicular  BC,  and  make  , 
it  equal  to  the  half  of  AB.  From  C 
as  a  center,  with  a  radius  equal  to  CB,  describe  a  circle. 
Draw  AC  cutting  the  circumference  in  D ;  and  make  AF 
equal  to  AD.  The  line  AB  will  be  d'jvided  in  the  point  F  in 
the  manner  required. 

For,  since  AB  is  a  perpendicular  to  the  radius  CB  at  its  ex- 
tremity, it  is  a  tangent  (Prop.  IX.,  B.  III.)  ;  and  if  we  pro- 
duce AC  to  E,  we  shall  have  AE  :  AB  :  :  AB  :  AD  (Prop. 
XXVIIL,  B.  IV.).  Therefore,  by  division  (Prop.  VII.,  B. 
£L),  AE— AB  :  AB  :  :  AB— AD  :  AD.  But,  by  construction, 
AB  is  equal  to  DE ;  and  therefore  AE — AB  is  equal  to  AD 
or  AF ;  and  AB— AD  is  equal  to  FB.  Hence  AF  :  AB  :  • 
FB  :  AD  or  AF ;  and,  consequently,  by  inversion  (Prop.  V 
B.  II.), 

AB  :  AF  :  :  AF  :  FB. 

Scholium.  The  line  AB  is  said  to  be  divided  in  extreme 
and  mean  ratio.  An  example  of  its  use  may  be  seen  in  Prop. 
V.,  Book  VI. 


PROBLEM    XXI. 

Through  a  given  point  in  a  given  angle,  to  draw  a  straight 
line  so  that  the  parts  included  between  the  point  and  the  sides 
of  the  angle,  may  be  equal. 

Let  A  be  the  given  point,  and  BCD  the 
given  angle  ;  it  is  required  to  draw  through 
A  a  line  BD,  so  that  BA  may  be  equal  to 
AD. 

Through  the  point  A  draw  AE  parallel  to 
BC  ;  and  take  DE  equal  to  CE.     Through 
ihe  points  D  and  A  draw  the  line  BAD ;  it  * 
ivilj  be  the  line  required. 


GEOMETRY 


For,  because  AE  is  parallel  to  BC  we  ihave  (Prop,  XVI, 
B.  IV.), 

DE  :  EC  :  :  DA  :  AB. 
But  DE  is  equal  to  EC  ;  therefore  DA  is  equal  to  AB. 


PROBLEM   XXII. 

To  describe  a  square  that  shall  be  equivalent  to  a  gnen 
parallelogram,  or  to  a  given  triangle. 

First.  Let  ABDC  be  the  given  paral-       *C  _  p 
telogram,  AB  its  base,  and  CE  its  altitude. 
Find  a  mean  proportional  between  AB  and 
CE  (Prob.  XIX.),  and  represent  it  by  X  ;     ^ 
the  square  described  on  X  will  be  equiva-  A    E  B 

lent  to  the  given  parallelogram  ABDC. 

For,  by  construction,  AB  :  X  :  :  X  :  CE  ;  hence  Xa  is  equaJ 
to  AB  X  CE  (Prop.  L,  Cor.,  B.  II.).  But  AB  X  CE  is  the 
measure  of  the  parallelogram  ;  and  X2  is  the  measure  of  the 
square.  Therefore  the  square  described  on  X  is  equivalent 
to  the  given  parallelogram  ABDC. 

Secondly.  Let  ABC  be  the  given  triangle, 
BC  its  base,  and  AD  its  altitude.  Find  a 
mean  proportional  between  BC  and  the  half 
of  AD,  and  represent  it  by  Y.  Then  will 
the  square  described  on  Y  be  equivalent  to 
the  triangle  ABC.  ' 

For,  by  construction,  BC  :  Y  :  :  Y  :  x  AD  ;  ence  a 
equivalent  to  BC  X  AD.  But  BC 


D 

hence  Y2 

AD  is  the  measure 

the  triangle  ABC  ;  therefore  the  square  described  on  Y 
eauivalent  to  the  triangle  ABC. 


is 


PROBLEM  XXIII. 

Upon  a  given  line,  to  construct  a  rectangle  equivalent  to  a 
given  rectangle. 

Let  AB  be  the  given  straight 
line,  and  CDFE  the  given  rect- 
angle. It  is  required  to  con- 
struct on  the  line  AB  a  rectan- 
gle equivalent  to  CDFE. 

Find    a    fourth    proportional  A 
Prob.  XVIII.)  to  the  three  lines  AB,  CD,  CE,  and  let  AG 
oe  that  fourth  proportional.     The  rectangle  constr  icted  on 
ihe  lines  AB,  AG  will  be  equivalent  to  CDFE. 


95 


For,  because  AB  :  CD  :  :  CE  :  AG,  by  Prop.  I.,  B.  II., 
ABxAG=CDxCE.  Therefore  the  rectangle  ABHG  is 
equivalent  to  the  rectangle  CDFE;  End  it.  is  constructed 
upon  the  given  line  AB. 


PROBLEM  XXIV. 

To  construct  a  triangle  which  shall  be  equivalent  to  a  given 
polygon. 

Let  ABODE  be  the  given  polygon ;  it 
s  required  to  construct  a  triangle  equiva- 
ent  to  it. 

Draw  the  diagonal  BD  cutting  off  the  E< 
triangle   BCD.      Through  the   point  C, 
draw  CF  parallel  to  DB,  meeting  AB 
produced  in  F.     Join  DF ;  and  the  poly- 
gon AFDE  will  be  equivalent  to  the  polygon  ABODE. 

For  the  triangles  BFD,  BCD,  being  upon  the  same  base 
BD,  and  between  the  same  parallels  BD,  FC,  are  equivalent. 
To  each  of  these  equals,  add  the  polygon  ABDE ;  then  will 
the  polygon  AFDE  be  equivalent  to  the  polygon  ABCDE ; 
that  is,  we  have  found  a  polygon  equivalent  to  the  given 
polygon,  and  having  the  number  of  its  sides  diminished  by 
one. 

In  the  same  manner,  a  polygon  may  be  found  equivalent 
to  AFDE,  and  having  the  number  of  its  sides  diminished  by 
one;   and,  by  continuing  the  process,  the  number  of  sides 
may  be  at  last  reduced  to  three,  and  a  triangle  be  thus  obtain 
ed  equivalent  to  the  given  polygon. 


PROBLEM  XXV. 

To  make  a  square  equivalent  to  the  sum  or  difference  of  twt 
(riven  squares. 

First.  To  make  a  square  equivalent  to  the  surr  of  twc 
given   squares.     Draw   two   indefinite   lines 
AB,  BC  at  right  angles  to  each  other.     Take 
AB  equal  to  the  side  of  one  of  the  given 

squares,  and    BC   equal  to  the   side  of  the     ^ 

other.     Join  AC  ;  it  will  be  the  side  of  the   A.  B 

required  square. 

For  the  triangle  ABC,  being  right-angle i  at  B,  the  squa-e 


*)6  GEOMETRY 

on  AC  will  be  equivalent  to  the  sum  of  the  squares  upon  AB 
and  BC  (Prop.  XL,  B.  IV.). 

Secondly.  To  make  a  square  equivalent  to  the  difference 
of  two  given  squares.  Draw  the  lines  AB,  BC  at  right  an 
gles  to  each  other ;  and  take  AB  equal  to  the  side  of  the  less 
square.  Then  from  A  as  a  center,  with*  a  radius  equal  to  the 
side  of  the  other  square,  describe  an  arc  intersecting  BC  in 
C ;  BC  will  be  the  side  of  the  square  required ;  because  the 
square  of  BC  is  equivalent  to  the  difference  of  the  squares  of 
AC  and  AB  (Prop.  XL,  Cor.  1,  B.  IV.). 

Scholium.  In  the  same  manner,  a  square  may  be  made 
equivalent  to  the  sum  of  three  or  more  given  squares ;  for 
the  same  construction  which  reduces  two  of  them  to  one 
will  reduce  three  of  them  to  two,  and  these  two  to  one. 


PROBLEM    XXVI. 

Upon  a  given  straight  line,  to  construct  a  polygon  simila 
to  a  given  polygon. 

Lot  ABODE  be  the  giv- 
en polygon,  and  FG  be 
the  given  straight  line ;  it 
/s  required  upon  the  line 
FG  to  construct  a  poly- 
gon similar  to  ABCDE. 

Draw  the  diagonals  BD,  }- 
BE.  At  the  point  F,  in  A 
the  straight  line  FG,  make  the  angle  GFK  equal  to  the  angle 
BAE ;  and  at  the  point  G  make  the  angle  FGK  equal  to  the 
angle  ABE.  The  lines  FK,  GK  will  intersect  in  K,  and 
FGK  will  be  a  triangle  similar  to  ABE.  In'  the  same  man- 
ner, on  GK  construct  the  triangle  GKI  similar  to  BED,  and 
on  GI  construct  the  triangle  GIH  similar  to  BDC.  The 
polygon  FGHIK  will  be  the  polygon  required.  For  these 
two  polygons  are  composed  of  the  same  number  of  triangles, 
which  are  similar  to  each  other,  and  similarly  situated  ;  there- 
fore the  polygons  are  similar  (Prop.  XXV.,  Ccr.,  B.  IV.) 


PROBLEM   XXVII. 

Given  the  area  of  a  rectangle,  and  the  sum  of  two  adjacent 
udes,  to  construct  the  rectangle. 

Let  AB  be  a  straight  line  equal  to  the  sum  of  the  sides  of 
he  reauired  rectangle. 


BOOK    V. 


-JO 


Upcn  AB  as  a  diameter,  describe  a 
semicircle.  At  the  point  A  erect  the 
perpendicular  AC,  and  make  it  equal  to- 

the  side  of  a  square  having  the  given 

area.     Through  C  draw  the  line  CD  par-  -A-  E   B 

allel  to  AB,  and  let  it  meet  the  circumference  in  D ;  and 
from  D  draw  DE  perpendicular  to  AB.  1  hen  will  AE  and 
EB  be  the  sides  of  the  rectangle  required. 

For,  by  Prop.  XXII.,  Cor.,  B.  IV.,  the  rectangle  AE  xEB 
is  equivalent  to  the  square  of  DE  or  CA,  which  is,  by  con- 
struction, equivalent  to  the  given  area.  Also,  the  sum  of  the 
sides  AE  and  EB  is  equal  to  the  given  line  AB. 

Scholium.  The  side  of  the  square  having  th%  given  .are  a, 
must  not  be  greater  than  the  half  of  AB  ;  for  in  that  case  the 
line  CD  would  not  meet  the  circumference  ADB. 


PROBLEM   XXVIII. 

Given  the  area  of  a  rectangle,  and  the  difference  of  two  ad- 
jacent sides,  to  construct  the  rectangle. 

Let  AB  be  a  straight  line  equal  to  the 
difference  of  the  sides  of  the  required  rect- 
angle. 

Upon  AB  as  a  diameter,  describe  a  cir- 
cle ;  and  at  the  extremity  of  the  diameter, 
draw  the  tangent  AC  equal  to  the  side  of 
a  square  having  the  given  area.  Through 
the  point  C  and  the  center  F  draw  the 
secant  CE ;  then  will  CD,  CE  be  the  adjacent  sides  of  the 
rectangle  required. 

For,  by  Prop.  XXVIIL,  B.  IV.,  the  rectangle  CDxCE  is 
equivalent  to  the  square  of  AC,  which  is,  by  construction, 
equivalent  to  the  given  area.  Also,  the  difference  of  the  lines 
CE,  CD  is  equal  to  DE  or  AB. 

E 


08  GEOMETRY. 


BOOK  VI. 

REGULAR  POLYGONS,  AND  THE  AREA  OF  THE  CIRCLE. 
Definition. 

A  regular  polygon  is  one  which  is  both  equiangular  and 
equilateral. 

An  equilateral  triangle  is  a  regular  polygon  of  three  sides  ; 
a  square  is  one  of  four. 

PROPOSITION  I.       THEOREM. 

Regular  polygons  of  the  same  number  of  sides  are  similar 
figures. 

Let  ABCDEF,  abcdef  be 
two  regular  polygons  of  the 
same  number  of  sides  ;  then 
will  they  be  similar  figures. 

For,  since  the  two  polygons 
have  the  same  number  of 
sides,  they  must  have  the 
same  number  of  angles.  Moreover,  the  sum  of  the  angles  of 
the  one  polygon  is  equal  to  the  sum  of  the  angles  of  the  other 
(Prop.  XXVIII.,  B.  I.) ;  and  since  the  polygons  are  each 
equiangular,  it  follows  that  the  angle  A  is  the  same  part  of 
the  sum  of  the  angles  A,  B,  C,  D,  E,  F,  that  the  angle  a  is 
of  the  sum  of  the  angles  a,  6,  c,  d,  e,  f.  Therefore  the  two 
angles  A  and  a  are  equal  to  each  other.  The  same  is  true 
of  the  angles  B  and  6,  C  and  c,  &c. 

Moreover,  since  the  polygons  are  regular,  the  sides  AB, 
BC,  CD,  &c.,  are  equal  to  each  other  (Def.)  ;  so,  also,  are  the 
sides  ab,  be,  cd,  &c.  Therefore  AB  :  ab  : :  BC  :  be  : :  CD  :  cd, 
&c.  Hence  the  two  polygons  have  their  angles  equal,  and 
their  homologous  sides  proportional ;  they  are  consequently 
similar  (Def.  3,  B.  IV.).  Therefore,  regular  polygons,  &c. 

Cor.  The  perimeters  of  two  regular  polygons  ot  the  same 
number  of  sides,  are  to  each  other  as  their  homologous  sides, 


BOOK    VI 


and  their  areas  are  us  the  squares  of  ,hose  aides  (Prop. 
XXVL,  B.  IV.). 

Scholium.     The  angles  of  a  regular  polygon   are  de'«r 
mined  by  the  number  of  its  sides. 


PROPOSITION  II.       THEOREM. 

A  circle  may  be  described  about  any  regular  polygon,  and 
another  may  be  inscribed  within  it. 

Let  ABCDEF  be  any  regular  polygon ; 
a  circle  may  be  described  about  it,  and 
another  may  be  inscribed  within  it. 

Bisect  the  angles  FAB,  ABC  by  the 
straight  lines  AO,  BO  ;  and  from  the  point 
O  in  which  they  meet,  draw  the  lines  OC. 
OD,  OE,  OF  to  the  other  angles  of  the 
polygon. 

Then,  because  in  the  triangles  OBA,  OBC,  AB  is,  by 
hypothesis,  equal  to  BC,  BO  is  common  to  the  two  triangles, 
and  the  included  angles  OBA,  OBC  are,  by  construction, 
equal  to  each  other ;  therefore  the  angle  OAB  is  equal  to  the 
ingle  OCB.  But  OAB  is,  by  construction,  the  half  of  FAB  ; 
ind  FAB  is,  by  hypothesis,  equal  to  DCB ;  therefore  OCB  is 
the  half  of  DCB ;  that  is,  the  angle  BCD  is  bisected  by  the 
line  OC.  In  the  same  mafiner  it  may  be  proved  that  the  an 
gles  CDE,  DEF,  EFA  are  bisected  by  the  straight  lines  OD 
OE,  OF. 

Now  because  the  angles  OAB,  OBA,  being  halves  of  equal 
angles,  are  equal  to  each  other,  OA  is  equal  to  OB  (Prop. 
XL,  B.  I.).  For  the  same  reason,  OC,  OD,  OE,  OF  are  each 
of  them  equal  to  OA.  Therefore  a  circumference  described 
from  the  center  O,  with  a  radius  equal  to  OA,  will  pass 
through  each  of  the  points  B,  C,  D,  E,  F,  and  be  described 
about  the  polygon. 

Secondly.  A  circle  may  be  inscribed  within  the  polygon 
ABCDEF.  For  the  sides  AB,  BC,  CD,  &c.,  are  equa. 
chords  of  the  same  circle ;  hence  they  are  equally  distant 
from  the  center  O  (Prop.  VIII.,  B.  III.)  ;  that  is,  the  perpen- 
diculars OG,  OH,  &c.,  are  all  equal  to  each  other.  There- 
fore, if  from  O  as  a  center,  with  a  radius  OG,  a  circumference 
be  described,  it  will  touch  the  side  BC  (Prop.  IX.,  B.  III.), 
and  each  of  the  other  sides  of  the  polygon  ;  hence  the  circle 
will  be  inscribed  within  the  polygon.  Therefore  a  circle 
may  be  described,  &c. 

Scholium  1.  In  regular  polygons,  the  center  of  the  inscribed 


IUO  GEOMETRY. 

and  circumscrioed  circles,  is  also  called  the  center  of  the  poly- 
gon; and  the  perpendicular  from  the  center  upon  one  of  the 
sides,  that  is,  the  radius  of  the  inscribed  circle,  is  called  the 
apothein  of  the  polygon. 

Since  all  the  chords  AB,  BC,  &c.,  are  equal,  the  angles  at 
the  center,  AOB,  BOG,  &c.,  are  equal ;  and  the  value  of  each 
may  be  found  by  dividing  four  right  angles  by  the  number 
of  sides  of  the  polygon. 

Scholium  2.  To  inscribe  a  regular  polygon  of  any  numbei 
of  sides  in  a  circle,  it  is  only  necessary  to  divide  the  circum- 
ference into  the  same  number  of  equal  parts  ;  for,  if  the 
arcs  are  equal,  the  chords  AB,  BC,  CD,  &c.,  will  be  equal. 
Hence  the  triangles  AOB,  BOC,  COD,  &c.,  will  also  be 
equal,  because  they  are  mutually  equilateral ;  therefore  all 
the  angles  ABC,  BCD,  CDE,  &c.,  will  be  equal,  and  the 
figure  ABCDEF  will  be  a  regular  polygon. 


PROPOSITION    III.       PROBLEM. 

* 

To  inscribe  a  square  in  a  given  circle. 

Let  ABCD  be  the  given  circle  ;  it  is  re- 
quired to  inscribe  a  square  in  it. 

Draw  two  diameters  AC,  BD  at  right 
angles  to  each  other ;  and  join  AB,  BC, 
CD,  DA.  Because  the  angles  AEB,  EEC, 
&c.,  are  equal,  the  chords  AB,  BC,  &c., 
are  also  equal.  And  because  the  angles 
ABC,  BCD,  &c.,  are  inscribed  in  semicir- 
cles, they  are  right  angles  (Prop.  XV.,  Cor.  2,  B.  III.). 
Therefore  ABCD  is  a  square,  and  it  is  inscribed  in  the  circle 
ABCD. 

Cor.  Since  the  triangle  AEB  is  right-angled  and  isosceles, 
we  have  the  proportion,  AB  :  AE  :  :  ^/2  :  1  (Prop.  XL,  Cor. 
3,  B.  IV.)  ;  therefore  the  side  of  the  inscribed  square  is  to  the 
"adius,  as  the  square  root  of  2  is  to  unity. 


PROPOSITION    IV.       THEOREM. 

The  side  of  a  regular  hexagon  is  equal  to  the  radius  of  th* 
circumscribed  circle. 

Let  ABCDEF  be  a  regular  hexagon  inscribed  in  a  circle 
whose  center  is  O;  then  any  side  as  AB  will  be  equal  to  the 
AO. 


BOOK    VI. 


101 


Draw  the  radius  BO.  Then  the  angle 
\OB  is  the  sixth  part  of  four  right  angles 
(Prop.  II.,  Sch.  1),  or  the  third  part  of  two 
right  angles.  Also,  because  the  three  an-  A|i 
gles  of  every  triangle  are  equal  to  two 
right  angles,  the  two  angles  OAB,  OBA 
are  together  equal  to  two  thirds  of  two 
right  angles ;  and  since  AO  is  equal  to  BO,  each  of  these  an« 
gles  is  one  third  of  two  right  angles.  Hence  the  triangle 
AOB  is  equiangular,  and  AB  is  equal  to  AO.  Therefore  the 
side  of  a  regular  hexagon,  &c. 

Cor.  To  inscribe  a  regular  hexagon  in  a  given  circle,  the 
radius  must  be  applied  six  times  upon  the  circumference. 
By  joining  the  alternate  angles  A,  C,  E,  an  equilateral  trian- 
gle will  be  inscribed  in  the  circle. 


PROPOSITION    V.       PROBLEM. 

To  inscribe  a  regular  decagon  in  a  given  circle. 

Let  ABF  be  the  given  circle ;  it  is  re- 
quired to  inscribe  in  it  a  regular  decagon. 

Take  C  the  center  of  the  circle ;  draw 
the  radius  AC,  and  divide  it  in  extreme 
and  mean  ratio  (Prob.  XX.,  B.  V.)  at 
the  point  D.  Make  the  chord  AB  equal 
to  CD  the  greater  segment ;  then  will 
AB  be  the  side  of  a  regular  decagon  in- 
scribed in  the  circle. 

Join  BC,  BD.     Then,  by  construction,  A 

AC  :  CD  :  :  CD  :  AD ;  but  AB  is  equal  to  CD ;  therefore 
AC  :  AB  :  :  AB  :  AD.  Hence  the  triangles  ACB,  ABD 
have  a  common  angle  A  included  between  proportional 
sides ;  they  are  therefore  similar  (Prop.  XX.,  B.  IV.)  And 
because  the  triangle  ACB  is  isosceles,  the  triangle  ABD  must 
also  be  isosceles,  and  AB  is  equal  to  BD.  But  AB  was  made 
equal  to^CD ;  hence  BD  is  equal  to  CD,  and  the  angle  DBC 
is  equal  to  the  angle  DCB.  Therefore  the  exterior  angle 
ADB,  which  is  equal  to  the  sum  of  DCB  and  DBC,  must  be 
double  of  DCB.  But  the  angle  ADB  is  equal  tc  DAB ;  there- 
fore each  of  the  angles  CAB,  CBA  is  double  of  the  angle 
ACB.  Hence  the  sum  of  the  three  angles  of  the  triangle 
ACB  is  five  times  the  angle  C.  But  these  three  angles  are 
equal  to  two  right  angles  (Prop.  XXVIL,  B.  I.) ;  therefore^ 
the  angle  C  is  the  fifth  part  of  two  right  angles,  or  the  tenth' 
part  of  four  right  angles.  Henc«  the  ar«  AB  is  one  ten/-J>  f 


102  GEOMETRY. 

the  circumference,  and  the  cnord  AB  is  the  side  of  a  regular 
decagon  inscribed  n  the  circle. 

Cor.  1.  By  joining  the  alternate  angles  of  the  regular  dec- 
agon, a  regular  pentagon  may  be  inscribed  in  the  circle. 

Cor.  2.  By  combining  this  Proposition  with  the  preceding, 
a  regular  pentedecagon  may  be  inscribed  in  a  circle. 

For,  let  AE  be  the  side  of  a  regular  hexagon  ;  then  the  arc 
AE  will  be  one  sixth  of  the  whole  circumference,  and  the  arc 
AB  one  tenth  of  the  whole  circumference.  Hence  the  arc 
BE  will  be  £ — TV  or  y1^,  and  the  chord  of  this  arc  will  be  the 
side  of  a  regular  pentedecagon. 

Scholium.  By  bisecting  the  arcs  subtended  by  the  sides  of 
any  polygon,  another  polygon  of  double  the  number  of  sides 
may  be  inscribed  in  a  circle.  Hence  the  square  will  enable 
us  to  inscribe  regular  polygons  of  8,  16,  32,  &c.,  sides;  the 
hexagon  will  enable  us  to  inscribe  polygons  of  12,  24,  &c., 
sides  ;  the  decagon  will  enable  us  to  inscribe  polygons  of 
20,  40,  &c.,  sides ;  and  the  pentedecagon,  polygons  of  30,  60, 
&c.,  sides. 

The  ancient  geometricians  were  unacquainted  with  any 
method  of  inscribing  in  a  circle,  regular  polygons  of  7,  9,  11, 
13,  14,  17,  &c.,  sides;  and  for  a  long  time  it  was  believed 
that  these  polygons  could  not  be  constructed  geometrically  ; 
but  Gauss,  a  German  mathematician,  has  shown  that  a  regu 
lar  polygon  of  17  sides  may  be  inscribed  in  a  circle,  by  em- 
Dloying  straight  lines  and  circles  only. 

PROPOSITION   VI.       PROBLEM. 

A  regular  polygon  inscribed'  in  a  circle  being  given,  to  dp 
scribe  a  similar  polygon  about  the  circle. 

Let  ABCDEF  be  a  regular  polygon 
inscribed  in  the  circle  ABD ;  it  is  re- 
quired to  describe  a  similar  polygon 
about  the  circle. 

Bisect  the  arc  AB  in  G,  and  through 
G  draw  the  tangent  LM.     Bisect  also 
the  arc  BC  in  H,  and  through  H  draw 
the  tangent  MN,  and  in  the  same  man- 
ner draw  tangents  to  the  middle  points 
of  the  arcs  CD,  DE,  &c,     These  tangents,  by  their  intersec- 
tions, will  form  a  circumsciibed  polygon  similar  to  the  one 
inscribed. 

Find  O  the  center  of  the  circle,  and  draw  the  radii  OG 
OH.  Then,  because  OG  is  perpendicular  to  the  tangent  LM 
(Prop.  IX.,  B.  III.),  and  also  to  the  chord  AB  (Prop.  VI 


BOOK    VI.  103 

Sch.,  B.  III.),  the  tangent  is  parallel  to  the  chord  (Prop.  XX., 
B.  I.).  In  the  same  manner  it  may  be  proved  that  the  other 
sides  of  the  circumscribed  polygon  are  parallel  to  the  sides 
of  the-  inscribed  polygon ;  and  therefore  the  angles  of  the 
circumscribed  polygon  are  equal  to  those  of  the  inscribed  one 
(Prop.  XXVI.,  B.  L). 

Since  the  arcs  BG,  BH  are  halves  of  the  equal  arcs  AGB, 
BHC,  they  are  equal  to  each  other ;  that  is,  the  vertex  B  is 
at  the  middle  point  of  the  arc  GBH.  Join  OM ;  the  line  OM 
will  pass  through  the  point  B.  For  the  right-angled  trian- 
gles OMH,  OMG  have  the  hypothenuse  OM  common,  and 
the  side  OH  equal  to  OG ;  therefore  the  angle  GOM  is  equal 
to  the  angle  HOM  (Prop.  XIX.,  B.  L),  and  the  line  OM  passes 
through  the  point  B,  the  middle  of  the  arc  GBH. 

Now  because  the  triangle  OAB  is  similar  to  the  triangle 
OLM,  and  the  triangle  OBC  to  the  triangle  OMN,  we  have 
the  proportions 

AB  :  LM  :  :  BO  :  MO ; 
also,  BC  :  MN  :  :  BO  :  MO ; 

therefore  (Prop.  IV.,  B.  II.), 

AB  :  LM  :  :  BC  :  MN. 

But  AB  is  equal  to  BC  ;  therefore  LM  is  equal  to  MN.  In 
the  same  manner,  it  may  be  proved  that  the  other  sides  of  the 
circumscribed  polygon  are  equal  to  each  other.  Hence  this 
polygon  is  regular,  and  similar  to  the  one  inscribed. 

Cor.  1.  Conversely,  if  the  circumscribed  polygon  is  given, 
and  it  is  required  to  form  the  similar  inscribed  one,  draw  the 
lines  OL,  OM,  ON,  &c.,  to  the  angles  of  the  polygon ;  these 
lines  will  meet  the  circumference  in  the  points  A,  B,  C,  &c. 
Join  these  points  by  the  lines  AB,  BC,  CD,  &c.,  and  a  similar 
polygon  will  be  inscribed  in  the  circle. 

Or  we  may  simply  join  the  points  of  contact  G,  H,  I,  &c. 
by  the  chords  GH,  HI,  &c.,  and  there  will  be  formed  an  in 
scribed  polygon  similar  to  the  circumscribed  one. 

Cor.  2.  Hence  we  can  circumscribe  about  a  circle,  any 
regular  polygon  which  can  be  inscribed  within  it,  and  con 
versely. 

Cor.  3.  A  side  of  the  circumscribed  polygon  MN  is  equa 
to  twice  MH,  or  MG  +  MH. 


PROPOSITION    VII.       THEOREM. 

The  area  of  a  regular  polygon  is  equivalent  to  the  product 
of  its  perimeter,  ~by  half  the  radius  of  the  inscribed  circle. 

Let  ABCDEF  be  a  regular  polygon,  and  G  the  center  01 


104 


GEOMETRY. 


the  inscribed  circle.  From  G  draw 
lines  to  all  the  angles  of  the  polygon. 
The  polygon  will  thus  be  divided  into 
as  many  triangles  as  it  has  sides ;  and 
the  common  altitude  of  these  triangles  A< 
is  GH,  the  radius  of  the  circle.  Now, 
.the  area  of  the  triangle  BGC  is  equal  to 
the  product  of  BC  by  the  half  of  GH 
(Prop.  VI.,  B.  IV.)  ;  and  so  of  all  the 
other  triangles  having  their  vertices  in  G.  Hence  the  sum 
of  all  the  triangles,  that  is,  the  surface  of  the  polygon,  is 
equivalent  to  the  product  of  the  sum  of  the  bases  AB,  BC. 
&c. ;  that  is,  the  perimeter  of  the  polygon,  multiplied  by  half 
of  GH,  or  half  the  radius  of  the  inscribed  circle.  Therefore, 
*he  area  of  a  regular  polygon,  &c. 


PROPOSITION    VIII.       THEOREM. 

The  perimeters  of  two  regular  polygons  of  the  same  numbei 
of  sides,  are  as  the  radii  of  the  inscribed  or  circumscribed  cir- 
cles, and  their  surfaces  are  as  the  squares  t>f  the  radii. 

Let  ABCDEF,  abcdef  be 
two  regular  polygons  of  the 
same  number  of  sides  ;  let  G 
and  g  be  the  centers  of  the 
circumscribed  circles ;  and 
let  GH,  gh  be  drawn  per- 
pendicular to  BC  and  be; 
then  will  the  perimeters  of  the  polygons  be  as  the  radii  BG. 
bg ;  and,  also,  as  GH,  gh,  the  radii  of  the  inscribed  circles. 

The  angle  BGC  is  equal  to  the  angle  bgc  (Prop.  II.,  Sch. 
1) ;  and  since  the  triangles  BGC,  bgc  are  isosceles,  they  are 
similar.  So,  also,  are  the  right-angled  triangles  BGH,  bgh  ; 
and,  consequently,  BC  :  be  :  :  BG  :  bg  :  :  GH  :  gh.  But  the 
perimeters  of  the  two  polygons  are  to  each  other  as  the  sides 
BC,  be  (Prop.  L,  Cor.)  ;  they  are,  therefore,  to  each  other  as 
the  radii  BG,  bg  of  the  circumscribed  circles ;  and  also  as  the 
radii  GH,  gh  of  the  inscribed  circles. 

The  surfaces  of  these  polygons  are  to  each  other  as  the 
squares  of  the  homologous  sides  BC,  be  (Prop.  L,  Cor.) ;  they 
are,  therefore,  as  the  squares  of  BG,  bg,  the  radii  of  the  cir 
cumscribed  circles  ;  or  as  the  squares  of  GH,  gh,  the  radii  of 
the  inscribed  circles. 


BOOK    Vr  103 


PROPOSITION    IX.       PROBLEM. 

The  surface  of  a  regular  inscribed  polygon,  and  tJiat  of  a 
similar  circumscribed  poly gon^being  given ;  to  find  the  surfaces 
of  regular  inscribed  and  circumscribed  polygons  having  double 
the  number  of  sides. 

Let  AB  be  a  side  of  the  given  in 
scribed  polygon ;  EF  parallel  to  AB,  a 
side  of  the  similar  circumscribed  poly- 
gon; and  C  the  center  of  the  circle. 
Draw  the  chord  AG,  and  it  will  be  the 
side  of  the  inscribed  polygon  having 
double  the  number  of  sides.  At  the 
points  A  and  B  draw  tangents,  meeting 
EF  in  the  points  H  and  I ;  then  will 
HI,  which  is  double  of  HG,  be  a  side  of 
the  similar  circumscribed  polygon  (Prop.  VI.,  Cor.  1).  Let 
p  represent  the  inscribed  polygon  whose  side  is  AB,  P  the 
corresponding  circumscribed  polygon ;  p1  the  inscribed  poly 
gon  having  double  the  number  of  sides,  P/  the  similar  cir- 
cumscribed polygon.  Then  it  is  plain  that  the  space  CAD  is 
the  same  part  of  p,  that  CEG  is  of  P ;  also,  GAG  of  pr,  and 
CAHG  of  P' ;  for  each  of  these  spaces  must  be  repeated  the 
same  number  of  times,  to  complete  the  polygons  to  which  they 
severally  belong. 

First.  The  triangles  ACD,  ACG,  whose  common  vertex  is 
A,  are  to  each  other  as  their  bases  CD,  CG ;  they  are  also  to 
each  other  as  the  polygons  p  and  p' ;  hence 
p  :  pf  :  :  CD  :  CG. 

Again,  the  triangles  CGA,  CGE,  whose  common  vertex  is 
G  are  to  each  other  as  their  bases  CA,  CE ;  they  are  also  to 
each  other  as  the  polygons  pr  and  P ;  hence 
pf  :  P  :  :  CA  :  CE. 

But  since  AD  is  parallel  to  EG,  we  have  CD  :  CG  :  :  CA 
CE  ;  therefore, 

that  is,  the  polygon  p'  is  a  mean  proportional  between  the 
two  given  polygons. 

Secondly.  The  triangles  CGH,  CHE,  having  the  common 
altitude  CG,  are  to  each  other  as  their  bases  GH,  HE.  But 
since  CH  bisects  the  angle  GCE,  we  have  (Prop.  XVII ,  B. 
IV.), 

GH:IIE::CG:CE:  :  CD    CA,  or  CG    :»:»'. 
Therefore,  CGH  :  CHE  :    p.p*; 


1 06  GEOMETRY. 

hence  (Prop.  VI.,  B.  II.) 

CGH  :  CGH  +  CHE,  or  CGE  :  .  p  :p+pf 
or  2CGH  :  CGE  :  :  2p  :  p+pt. 

But  2CGH,  or  CGHA  :  CGE  :  :  P'     P. 

Therefore,  F  :  P  ;  :  2p  :  p+p' ;  whence  P/- 

that  is,  the  polygon  P'  is  found  by  dividing  twice  the  product  oj 
the  two  given  polygons  by  the  sum  of  the  two  inscribed  polygons 
Hence,  by  means  of  the  polygons  p  and  P,  it  is  easy  to  find 
the  polygons  p1  and  P'  having  double  the  number  of  sides. 


PROPOSITION    X.       THEOREM. 

A  circle  being  given,  two  similar  polygons  can  always  be 
found,  the  one  described  about  the  circle,  and  the  other  inscribed 
in  it,  which  shall  differ  from  each  other  by  less  than  any  as- 
signable surface. 

Let  ACD  be  the  given  circle,  and  the  square  of  X  any 
given  surface ;  a  polygon  can  be  inscribed  in  the  circle 
ACD,  and  a  similar  polygon  be  described  about  it,  such 
that  the  difference  between  them  shall  be  less  than  the 
square  of  X. 

Bisect  AC  a  fourth  part  of  the  circumference,  then  bisect ' 
the  half  of  this  fourth,  and  so  continue  the  bisection,  until  an 
arc  is  found  whose  chord  AB  is  less  than  X.  As  this  arc 
must  be  contained  a  certain  number  of  times  exactly  in  the 
whole  circumference,  if  we  apply  chords  AB,  BC,  &c.,  each 
equal  to  AB,  the  last  will  tarminate  at  A,  and  a  regular 
polygon  ABCD,  &c.,  will  be  inscribed  in  the  circle. 

Next  describe  a  similar  polygon  about  the  circle  (Prop. 
VI.)  ;  the  difference  of  these  two  polygons  will  be  less  than 
the  square  of  X. 

Find  the  center  G,  and  draw  the 
diameter  AD.  Let  Et  be  a  side 
of  the  circumscribed  polygon ;  and 
join  EG,  FG.  These  lines  will  pass 
through  the  points  A  and  B,  as  was 
shown  in  Prop.  VI.  Draw  GH  to 
the  point  of  contact  H  ;  it  will  bisect 
AB  in  I,  and  be  perpendicular  to  it 
(Prop.  VI.,  Sch.,  B.  III.).  Join,  also, 
BD. 

Let  P  represent  the  circumscribed  polygon,  and  p  the  in- 
scribed polygon.  Then,  because  the  polygons  are  similar, 
they  are  as  the  squares  of  the  homologous  sides  EF  and  AB 


BOOK    VI.  107 

(Prop.  XXVL,  B.  IV.) ;  that  is,  because  the  triangles  EFG 
ABG  are  similar,  as  the  square  of  EG  to  the  square  of  AG 
-hat  is,  of  HG. 

Again,  the  in  angles  EHG,  ABD,  having  their  sides  paral 
.el  to  each  other,  are  similar ;  and,  therefore, 

EG  :  HG  :  :  AD  :  BD. 

But  the  polygon  P  is  to  the  polygon  p  as  the  square  of  EG 
to  the  square  of  HG ; 
hence  P  :  p  :  :  AD2  :  BD2, 

and,  by  division,  P  :  P— p  : :  AD2  :  AD2— BDS,  or  AB2. 
But  the  square  of  AD  is  greater  than  a  regular  p.a.ygon  of 
eight  sides  described  about  the  circle,  because  it  contains 
that  polygon ;  and  for  the  same  reason,  the  polygon  of  eight 
sides  is  greater  than  the  polygon  of  sixteen,  and  so  on. 
Therefore  P  is  less  than  the  square  of  AD  ;  and,  consequent- 
ly (Def.  2,  B.  II.),  P— p  is  less  than  the  square  of  AB  ;  that  is, 
less  than  the  given  square  on  X.-  Hence,  the  difference  of 
the  two  polygons  is  less  than  the  given  surface. 

Cor.  Since  the  circle  can  not  be  less  than  any  inscribed 
polygon,  nor  greater  than  any  circumscribed  one,  it  follows 
that  a  polygon  may  be  inscribed  in  a  circle,  and  another  de- 
scribed about  it,  each  of  which  shall  differ  from  the  circle  by 
less  than  any  assignable  surface. 


PROPOSITION    XI.       PROBLEM. 

To  find  the  area  of  a  circle  whose  radius  is  unity-. 

If  the  radius  of  a  circle  be  unity,  the  diameter  will  be  rep 
resented  by  2,  and  the  area  of  the  circumscribed  square  wiL 
oe  4 ;  while  that  of  the  inscribed  square,  being  half  the  cir- 
cumscribed, is  2.  Now,  according  to  Prop.  IX.,  the  surface 
of  the  inscribed  octagon,  is  a  mean  proportional  between  the 
two  squares  p  and  P,  so  that  pf=^/8= 2.82843.  Also,  the 

circumscribed  octagon  P'=-A-  = =3.31371.     Hav- 

p+p1     2+v/8 

ing  thus  obtained  the  inscribed  and  circumscribed  octagons, 
we  may  in  the  same  way  determine  the  polygons  having 
twice  the  number  cf  sides.  We  must  put  p  =  2.82843,  and 
P  =  3.31371,  and  we  shall  have  p'=  V~pP  =  3.06 147  ;  and 

P/=-^-7=3.18260.     These  polygons  of  16  sides  will  furnish 

us  those  of  32 ;  and  thus  we  may  j  *oceed,  until  there  is  no 
difference  between  the  inscribed  and  circumscribed  polygons, 
at  least  for  anv  number  of  decimal  pi  aces  which  may  b<*  de- 


10W  GEOMETRY 

sired.     The  following  table  gives  the  results  of  this  compute 
tion  for  five  decimal  places  : 

Number  of  Sides.  Inscribed  Polygon.  Circumscribed  Polygon 

4  2.00000  4.00000 

8  2.82843  3.31371 

16  3.06147  3.18260 

32  3.12145  3.15172 

64  3.13655  3.14412 

128  3.14033  3.14222 

256  3.14128  3.14175 

512  3.14151  3.14163 

1024  3.14157  3.14160 

2048  3.14159  3.14159 

Now  as  the  inscribed  polygon  can  not  be  greater  than  the 
circle,  and  the  circumscribed  polygon  can  not  be  less  than 
the  circle,  it  is  plain  that  3.14159  must  express  the  area  of  a 
circle,  whose  radius  is  unity,  correct  to  five  decimal  places. 

After  three  bisections  of  a  quadrant  of  a  circle,  we  obtain 
the  inscribed  polygon  of  32  sides,  which  differs  from  the  cor- 
responding circumscribed  polygon,  only  in  the  second  decimal 
place.  After  five  bisections,  we  obtain  polygons  of  128  sides, 
which  differ  only  in  the  third  decimal  place  ;  after  nine  bisec- 
tions, they  agree  to  five  decimal  places,  but  differ  in  the  sixth 
place ;  after  eighteen  bisections,  they  agree  to  ten  decimal 
places  ;  and  thus,  by  continually  bisecting  the  arcs  subtended 
by  the  sides  of  the  polygon,  new  polygons  are  formed,  both 
inscribed  and  circumscribed,  which  agree  to  a  greater  num- 
ber of  decimal  places.  Vieta,  by  means  of  inscribed  and  cir- 
cumscribed polygons,  carried  the  approximation  to  ten  places 
of  figures ;  Van  Ceulen  carried  it  to  36  places ;  Sharp  com- 
puted the  area  to  72  places ;  De  Lagny  to  128  places ;  and 
Dr.  Clausen  has  carried  the  computation  to  250  places  of  de- 
cimals. 

By  continuing  this'  process  of  bisection,  the  difference  be- 
tween the  inscribed  and  circumscribed  polygons  may  be 
made  less  than  any  quantity  we  can  assign,  however  small. 
The  number  of  sides  of  such  a  polygon  will  be  indefinitely 
great ;  and  hence  a  regular  polygon  of  an  infinite  number  of 
sides,  is  said  to  be  ultimately  equal  to  the  circle.  Henceforth, 
we  shall  therefore  regard  the  circle  as  «i  regular  polygon  of 
an  infinite  number  of  sides. 


BOOK    VI. 


PROPOSITION   XII.       THEOREM. 


The  ar*2  of  a  circle  is  equal  to  the  product  of  its  circuits 
\erence  by  half  the  radius. 

Let  ABE  be  a  circle  whose  center  is  C 
and  radius  CA  ;  the  area  of  the  circle  is 
*qual  to  the  product  of  its  circumference  by 
half  of  CA. 

Inscribe  in  the  Circle  any  regular  polygon, 
and  from  the  center  draw  CD  perpendicular 
to  one  of  the  sides.  The  area  of  the  poly- 
gon will  be  equal  to  its  perimeter  multiplied  A 
by  half  of  CD  (Prop.  VII.).  Conceive  the  number  of  sides 
of  the  polygon  to  be  indefinitely  increased,  by  continually 
bisecting  the  arcs  subtended  by  the  sides  ;  its  perimeter 
will  ultimately  coincide  with  the  circumference  of  the  circle 
the  perpendicular  CD  will  become  equal  to  the  radius  CA 
and  the  area  of  the  polygon  to  the  area  of  the  circle  (Prop 
XL).  Consequently,  the  area  of  the  circle  is  equal  to  tb<~ 
product  of  its  circumference  by  half  the  radius. 

Cor.   The  area  of  a  sector  is  equal  to  the  product  of  its  arc 
by  half  its  radius. 

For  the  sector  ACB  is  to  the  whole  circle 
ABD,  as  the  arc  AEB  is  to  the  whole  cir- 
cumference ABD  (Prop.  XIV.;   Cor.   2,  B. 
III.)  ;   or,  since  magnitudes  have   the   same 
ratio  which  their  equimultiples  have  (Prop. 
VIIL,  B.  II.),  as  the  arc  AEB  XI  AC  is  to  the 
circumference  ABD  X  |AC.  ;    But  this  last  ex-      "V*  .......  .*•' 

pression  is  equal  to  the  area  of  the  circle  ;       *     I) 
therefore  the  area  of  the  sector  ACB  is  equal  to  the  product 
of  its  arc  AEB  by  half  of  AC. 


PROPOSITION   XIII.       THEOREM. 

The  circumferences  of  circles  are  to  each  other  as  their  radii, 
and  their  areas  are  as  the  squares  of  their  radii. 

Let  R  and  r  denote  the  radii  of  two  circles  ;  C  and  c  their 
circumferences  ;  A  and  a  their  areas  ;  then  we  shall  have 

C  :  c  :  :  R   :  r. 
and  A  :  a  :  :  R3  :  rtt 

Inscribe  within  the  circles,  two  regular  polygons  having 


110  GEOMETK    . 

the  same  number  of  sides.  Now  wnatever  be  tne  numbei 
of  sides  of  the  polygons,  their  perimeters  will  be  to  each  other 
as  the  radii  of  the  circumscribed  circles  (Prop.  VIII.).  Con- 
ceive the  arcs  subtended  by  the  sides  of  the  polygons  to  be 
continually  bisected,  until  the  number  of  sides  of  the  polygons 
becomes  indefinitely  great,  the  perimeters  of  the  polygons  will 
ultimately  become  equal  to  the  circumferences  of  the  circles, 
find  we  shall  have 

C  :  c  :  :  R  :  r. 

Again,  the  areas  of  the  polygons  are  to  each  other  as  the 
squares  of  the  radii  of  the  circumscribed  circles  (Prop.  VIIL). 
But  when  the  number  of  sides  of  the  polygons  is  indefinitely 
increased,  the  areas  of  the  polygons  become  equal  to  the 
ureas  of  the  circles,  and  we  shall  have 
A  :  a  :  :  Ra  :  r\ 

Cor.  1.  Similar  arcs  are  to  each  other  as  their  radii;  and 
similar  sectors  are  as  the  squares  of  their  radii. 

For  since  the  arcs  AB,  ab  are 
«imilar,  the  angle  C  is  equal  to  the 
single  c  (Def.  5,  B.  IV.).  But  the 
ingle  C  is  to  four  right  angles,  as 
tfie  arc  AB  is  to  the  whole  circum- 
ference described  with  the  radius 
AC  (Prop.  XIV.,  B.  III.) ;  and  the 
angle  c  is  to  four  right  angles,  as  the  arc  ab  is  to  the  circum- 
ference described  with  the  radius  ac.  Therefore  the  arcs 

AB,  ab  are  to  each  other  as  the  circumferences  of  which  they 
form  a  part.     But  these  circumferences  are  to  each  other  as 

AC,  ac ;  therefore, 

Arc  AB  :  arc  ab  :  :  AC  :  ac. 

For  the  same  reason,  the  sectors  ACB,  acb  are  as  the  en 
tire  circles  to  which  they  belong ;  and  these  are  as  the  squares 
of  their  radii ;  therefore, 

Sector  ACB  :  sector  acb  :  :  AC2  :  ac\ 

Cor.  2.  Let  n  represent  the  circumference  of  a  circle  whose 
diameter  is  unity ;  also,  let  D  represent  the  diameter,  R  the 
radius,  and  C  the  circumference  of  any  other  circle ;  then, 
since  the  circumferences  of  circles  are  to  each  other  as  theii 
diameters, 

1  :  rr  :  :  2R  :  C ; 

therefore,  C  =  2nR = nD  • 

that  is,  the  circumference  of  a  circle  is  equal  to  the  product  of 
its  diameter  by  the  constant  number  n. 

Cor.  3.  According  to  Prop.  XII.,  the  area  of  a  circle  is 
equal  to  the  product  of  its  circumference  by  half  the  radius 

If  we  put  A  to  represent  the  area  of  a  circle,  then 


BOOK     /i.  Ill 

that  is,  the  ar-a  of  a  circle  is  equal  to  the  product  of  the  square 
of  its  radius  ~y  the  constant  number  TT. 

Cor.  4,  When  R  is  equal  to  unity,  we  have  A=TT  ;  that  is, 
TT  is  equal  to  the  area  of  a  circle  whose  radius  is  unity.  Ac- 
cording to  Prop.  XL,  TT  is  therefore  equal  to  3.14159  nearly 
This  number  is  represented  by  TT,  because  it  is  the  first  letter 
of  the  Greek  word  which  signifies  circumference. 


112  'JEOMETRY. 


SOLID  GEOMETRY. 

BOOK  VII. 

PLANES  AND  SOLID  ANGLES 
Definitions. 

1.  A  STRAIGHT  line  is  perpendicular  to  a 
plane,  when  it  is  perpendicular  to  every 
straight  line  which  it  meets  in  that  plane. 

Conversely,  the  plane  in  this  case  is  per 
pendicular  to  the  line. 

The  foot  of  the  perpendicular,  is  the 
point  in  which  it  meets  the  plane. 

2.  A  line  is  parallel  to  a  plane,  when  it  can  not  meet  the 
plane,  though  produced  ever  so  far. 

Conversely,  the  plane  in  this  case  is  parallel  to  the  line. 

3.  Two  planes  are  parallel  to  each  other,  when  they  can 
not  meet,  though  produced  ever  so  far. 

4.  The  angle  contained  by  two  planes  which  cut  each  othe«-f 
.3  the  angle  contained  by  two  lines  drawn 

from  any  point  in  the  line  of  their  common 
section,  at  right  angles  to  that  line,  one  in 
each  of  the  planes. 

This  angle  may  be  acute,  right,  or  obtuse. 

If  it  is  a  right  angle,  the  two  planes  are 
perpendicular  to  each  other. 

5.  A  solid  angle  is  the  angular  space  con- 
tained by  more  than  two  planes  which  meet  at 
the  same  point. 


PROPOSITION  I.       THEOREM 


One  part  of  a  straight  line  can  not  be  in  a  plane,  and  anothet 
pert  without  it. 

For  from  the  definition  of  a  plane  (Def.  6,  B.  I.),  wlun  o 


BOOK    VII.  113 


straight  line  has  two  points  common  with-  a  plane    it  lies 
wholly  in  that  plane. 

Scholium.  To  discover  whether  a  surface  is  plane,  we  ap 
ply  a  straight  line  in  different  directions  to  this  surface,  and 
Kee  if  it  touches  throughout  its  whole  extent. 


PROPOSITION  II.       THEOREM. 

Any  two  straight  lines  which  cut  each  other,  are  in  one  plane, 
and  determine  its  position. 

Let  the  two  straight  lines  AB,  BC  cut 
each  other  in  B;  then  will  AB,  BC  be  in 
the  same  plane. 

Conceive  a  plane  to  pass  through  the 
straight  line  BC,  and  let  this  plane  be  turned  Z 
about  BC,  until  it  pass  through  the  point  A.  B 
Then,  because  the  points  A  and  B  are  situated  in  this  plane 
the  straight  line  AB  lies  in  it  (Def.  6,  B.  I.).  Hence  the  posi- 
tion of  the  plane  is  determined  by  the  condition  of  its  con- 
taining the  two  lines  AB,  BC.  Therefore,  any  two  straight 
lines,  ccc. 

Cor.  1.  A  triangle  ABC,  or  three  points  A,  B,  C,  not  in  the 
same  straight  line,  determine  the  position  of  a  plane. 

Cor.  2.  Two  parallel  lines   AB,  CD  N 

determine  the  position  of  a  plane.     For  ^          \E g 

if  the  line  EF  be  drawn,  the  plane  of  \ 

the  two  straight  lines  AE,  EF  will  be  G -4 D 

the  same  as  that  of  the  parallels  AB, 

CD ;  and  it  has  already  been  proved  that  two  straight  lines 

which  cut  each  other,  determine  the  position  of  a  plane 


PROPOSITION  III.       THEOREM. 

If  two  planes  cut  each  other,  their   common  section  is  a 
M'J  aight  line. 

Let  the  two  planes  AB,  CD  cut  each 
other,  and  let  E.  F  be  two  points  in  their 
common  section.  From  E  to  F  draw  the 
straight  line  EF.  Then,  since  the  points  E 
and  F  are  in  the  plane  AB,  the  straight  line 
EF  which  joins  them,  must  lie  wholly  in 
that  plane  (Def.  6,  B.  L).  For  the  same 
reason,  EF  must  lie  wholly  in  the  plane 


114  GEOMETRY. 


UD.  Therefore  the  straight  line  EF  is  common  to  the  two 
planes  AB,  CD ;  that  is,  it  is  their  common  section.  Hence, 
if  two  planes,  &c. 


PROPOSITION  IV.       THEOREM. 


If  a  straight  line  be  perpendicular  to  each  of  two  straight 
Lines  at  their  point  of  intersection,  it  will  be  perpendicular  to 
the  plane  in  which  these  lines  are. 

Let  the -straight  line  AB  be  perpen- 
dicular to  each  of  the  straight  lines 
CD,  EF  which  intersect  at  B ;  AB  will 
also  be  perpendicular  to  the  plane  MN 
which  passes  through  these  lines. 

Through  B  draw  any  line  BG,  in  the 
plane  MN ;  let  G  be  any  point  of  this 
line,  and  through  G  draw  DGF,  so  that 
DG  shall  be  equal  to  GF  (Prob.  XXL, 
B.  V.).  Join  AD,  AG,  and  AF. 

Then,  since  the  base  DF  of  the  triangle  DBF  is  bisected 
in  G,  we  shall  have  (Prop.  XIV.,  B.  IV.), 

BD2+BF2=2BG2+2GF2. 
Also,  in  the  triangle  DAF, 

AD2+AF2=2AG2+2GF2. 
Subtracting  the  first  equation  from  the  second,  we  have 

AD2— BD2+AF2— BF2=2AG2— 2BG2. 
But,  because  ABD  is  a  right-angled  triangle, 

AD2-BD2=AB2; 
and,  because  ABF  is  a  right-angled  triangle, 

AP-BF2=AB*. 
Therefore,  substituting  these  values  in  the  former  equation, 

AB2+AB2=2AG2-2BG2 ; 
whence  AB2 = AG2  -  BG8, 

or  AG2=AB2+BG2. 

Wherefore  ABG  is  a  right  angle  (Prop.  XIIL,  Sch.,  B.  IV.) 
that  is,  AB  is  perpendicular  to  the  straight  line  BG.  In  like 
manner,  it  may  be  proved  that  AB  is  perpendicular  to  any 
other  strai^*  Tine  passing  through  B  in  the  plane  MN ;  hence 
it  is  perpewrflcular  to  the  plane  MN  (Def.  1).  Therefore,  if 
a  straight  line,  &c. 

Scholium.  Hence  it  appears  not  only  that  a  straight  line 
may  be  perpendicular  to  every  straight  line  which  passes 
through  its  foot  in  a  plane,  but  that  it  always  must  be  so 
whenever  it  is  perpendicular  to  two  lines  in  the  plane,  wWh 
shows  that  the  first  defini£km  involves  no  impossibility. 


BOOK    VII.  115 

Cor.  1  The  perpendicular  AB  is  shorter  than  any  oblique 
ine  AD ;  it  therefore  measures  the  true  distance  of  the  point 
A  from  the  plane  MN. 

Cor.  2.  Through  a  given  point  B  in  a  plane,  only  one  per- 
pendicular can  be  drawn  to  this  plane.  For,  if  there  could 
be  two  perpendiculars,  suppose  a  plane  to  pass  through  them, 
whose  intersection  with  the  plane  MN  is  BG;  then  these 
two  perpendiculars  would  both  be  at  right  angles  to  the  line 
BG,  at  the  same  point  and  in  the  same  plane,  which  is  im- 
possible (Prop.  XVL,  Cor.,  B.  L). 

It  is  also  impossible,  from  a  given  point  without  a  plane,  to 
let  fall  two  perpendiculars  upon  the  plane.  For,  suppose  AB, 
AG  to  be  two  such  perpendiculars;  then  the  triangle  ABG 
will  have  two  right  angles,  which  is  impossible  (Prop.  XXVIL, 
Cor.  3,  B.  L). 


PROPOSITION   V.       THEOREM. 

Oblique  lines  drawn  from  a  point  to  a  plane,  at  equal  dis- 
tances from  the  perpendicular,  are  equal ;  and  of  two  oblique 
lines  unequally  distant  from  the  perpendicular,  the  more  remote 
is  the  longer. 

Let  the  straight  line  AB  be 
drawn  perpendicular  to  the  plane 
MN;  and  let  AC,  AD,  'AE  be  ob- 
lique lines  drawn  from  the  point  A., 
equally  distant  from  the  perpendic- 
ular ;  also,  let  AF  be  more  remote 
from  the  perpendicular  than  AE ; 
then  will  the  lines  AC,  AD,  AE  all 
be  equal  to  each  other,  and  AF  be 
longer  than  AE. 

For,  since  the  angles  ABC,  ABD,  ABE  are  right  angles 
and  BC,  BD,  BE  are  equal,  the  triangles  ABC,  ABD,  ABE 
have  two  sides  and  the  included  angle  equal ;  therefore  the 
third  sides  AC,  AD,  AE  are  equal  to  each  other. 

So,  also,  since  the  distance  BF  is  greater  than  BE,  it  is 
plain  that  the  oblique  line  AF  is  longer  than  AE  (Prop.  XVII., 
B.  L). 

Cor.  All  the  equal  oblique  lines  AC,  AD,  AE,  &c.,  termi- 
nate in  the  circumference  CDE,  which  is  described  from  B, 
the  foot  of  the  perpendicular,  as  a  center. 

If,  then,  it  is  required  to  draw  a  straight  line  perpendiculai 
to  the  plane  MN,  from  a  point  A  without  it,  take  three  points 
in  the  plane  C,  D,  E,  equally  distant  from  A,  and  find  B  th« 


110  GEOMETRY. 

center  of  the  circle  which  passes  througn  these  points.     Join 
AB,  and  it  will  be  the  perpendicular  required. 

Scholium.  The  angle  AEB  is  called  the  inclination  of  t/ia 
line  AE  to  the  plane  MN.  All  the  lines  AC,  AD,  AE,~&cM 
which  are  equally  distant  from  the  perpendicular,  have  the 
same  inclination  to  the  plane;  because  all  the  angles  ACB 
ADB,  AEB,  &c.,  are  equal. 

PROPOSITION  VI.       THEOREM. 

If  a  straight  line  is  perpendicular  to  a  plane,  every  plane 
which  passes  through  that  line,  is  perpendicular  to  the  first- 
mentioned  plane. 

Let  the  straight  line  AB  be  perpen- 
dicular to  the  plane  MN ;  then  will 
every  plane  which  passes  through  AB 
be  perpendicular  to  the  plane  MN. 

Suppose  any  plane,  as  AE,  to  pass 
through  AB,  and  let  EF  be  the  common 
section  of  the  planes  AE,  MN.  In  the 
plane  MN,  through  the  point  B,  draw 
CD  perpendicular  to  the  common  sec- 
tion EF.  Then,  since  the  line  AB  is  perpendicular  to  the 
plane  MN,  it  must  be  perpendicular  to  each  of  the  two 
straight  lines  CD,  EF  (Def.  1).  But  the  angle  ABD,  formed 
by  the  two  perpendiculars  BA,  BD,  to  the  common  section 
EF,  measures  the  angle  of  the  two  planes  AE,  MN  (Def.  4) ; 
and  since  this  is  a  right  angle,  the  two  planes  must  be  per- 
pendicular to  each  other.  Therefore,  if  a  straight  line,  &c. 

Scholium.  When  three  straight  lines,  as  AB,  CD,  EF,  are 
perpendicular  to  each  other,  each  of  these  lines  is  perpen- 
dicular to  the  plane  of  the  other  two,  and  the  three  planes 
are  perpendicular  to  each  other. 


PROPOSITION    VII        THEOREM. 

If  two  planes  are  perpendicular  to  each  other,  a  straight  line 
drawn  in  one  of  them  perpendicular  to  their  common  section* 
will  be  perpendicular  to  the  other  plane. 

Let  the  plane  AE  be  perpendicular  to  the  plane  MN,  and 
let  the  line  AB  be  drawn  in  the  plane  AE  perpendicular  to 
the  common  section  EF ;  then  will  AB  be  perpendicular  to 
the  plane  MN. 


BOOK    Vif. 


117 


For  in  the  plane  MN,  draw  CD 
tnrough  the  point  B  perpendicular  to 
EF.  Then,  because  the  planes  AE  and 
MN  are  perpendicular,  the  angle  ABD 
is  a  right  angle.  Hence  the  line  AB  is 
perpendicular  to  the  two  straight  lines 
CD,  EF  at  their  point  of  intersection ; 
it  is  consequently  perpendicular  to  their 
plane  MN  (Prop.  IV.).  Therefore,  if 
two  planes,  &c. 

Cor.  If  the  plane  AE  is  perpendicular  to  the  plane  MN, 
and  if  from  any  point  B,  in  their  common  section,  we  erect  a 
perpendicular  to  the  plane  MN,  this  perpendicular  will  be  in 
the  plane  AE.  For  if  not,  then  we  may  draw  from  the  same 
point,  a  straight  line  AB  in  the  plane  AE  perpendicular  to 
EF,  and  this  line,  according  to  the  Proposition,  will  be  per- 
pendicular to  the  plane  MN.  Therefore  there  would  be  two 
perpendiculars  to  the  plane  MN,  drawn  from  the  same  point, 
which  is  impossible  (Prop.  IV.,  Cor.  2). 


PROPOSITION   VIII.       THEOREM. 

If  two  planes,  which  cut  one  another,  are  each  of  them  per- 
pendicular to  a  third  plane,  their  common  section  is  perpen* 
dicular  to  the  same  plane. 

Let  the  two  planes  AE,  AD  be  each 
of  them  perpendicular  to  a  third  plane 
MN,  and  let  AB  be  the  common  sec- 
tion of  the  first  two  planes ;  then  will 
AB  be  perpendicular  to  the  plane  MN. 

For,  from  the  point  B,  erect  a  per- 
pendicular to  the  plane  MN.  Then,  by 
the  Corollary  of  the  last  Proposition, 
this  line  must  be  situated  both  in  the 
plane  AD  and  in  the  plane  AE ;  hence  it  is  their  common 
f  ection  AB.  Therefore,  if  two  planes,  &c. 


PROPOSITION    IX.       THEOREM. 

Two  straight  lines  which  are  perpendicular  to  the  same  plane, 
ire  i'-'J-iJlel  to  each  other. 


Let  the  two  sti.  :grht  lines  AB,  CD  be  each  of  them  perpen- 
dicular to  the'same  plane  MN  ;  then  will  AB  be  parallel  to  CD 


118  GEOMETRY. 

In  the  plane  MN,  draw  the  straight 
line  BD  joining  the  points  B  and  D. 
Through  the  lines  AB,  BD  pass  the 
plane  EF;  it  will  be  perpendicular  to 
the  plane  MN  (Prop.  VI.) ;  also,  the 
line  CD  will  lie  in  this  plane,  because  it 
is  perpendicular  to  MN  (Prop.  VII., 
Cor.).  Now,  because  AB  and  CD  are 
both  perpendicular  to  the  plane  MN, 
they  are  perpendicular  to  the  line  BD  in  that  plane  ;  and  since 
AB,  CD  are  both  perpendicular  to  the  same  line  BD,  and  lie 
m  the  same  plane,  they  are  parallel  to  each  other  (Prop. 
XX.,  B.  I.).  Therefore,  two  straight  lines,  &c. 

Cor.  1.  If  one  of  two  parallel  lines  be  perpendicular  to  a 
plane,  the  other  will  be  perpendicular  to  the  same  plane.  If 
AB  is  perpendicular  to  the  plane  MN,  then  (Prop.  VI.)  the 
plane  EF  will  be  perpendicular  to  MN.  Also,  AB  is  per- 
pendicular to  BD ;  and  if  CD  is  parallel  to  AB,  it  will  be 
perpendicular  to  BD,  and  therefore  (Prop.  VII.)  it  is  perpen- 
dicular to  the  plane  MN. 

Cor.  2.  Two  straight  lines,  parallel  to  a  third,  are  parallel 
to  each  other.  For,  suppose  a  plane  to  be  drawn  perpen- 
dicular to  any  one  of  them ;  then  the  other  two,  being  paral- 
lel to  the  first,  will  be  perpendicular  to  the  same  plane,  by 
the  preceding  Corollary ;  hence,  by  the  Proposition,  they  wil- 
be  parallel  to  each  other. 

The  three  straight  lines  are  supposed  not  to  be  in  the  same 
plane ;  for  in  this  case  the  Proposition  has  been  already  de 
monstrated 


PROPOSITION    X.       THEOREM. 

If  a  straight  tine,  without  a  given  plane,  be  parallel  to  a 
straight  line  in  the  plane,  it  will  be  parallel  to  the  plane. 

Let  the  straight  line  AB  be  parallel  A  U 

to  the  straight  line  CD,  in  the  plane  /~~ 

MN;   then   will  it  be  parallel   to  the  M^jL  -^       /, 
plane  MN.  W&-  l\ 

Through  the  parallels  AB,  CD  sup-       V/fp         ~- ly  \ 
pose  a  plane  ABDC  to  pass.     If  the  line        \__  __\ 

AB  can  meet  the  plane  MN,  it  must  N 

meet  it  in  some  point  of  the  line  CD,  which  is  the  common 
intersection  of  the  two  planes.     But  AB  can  not  meet  CD 
since  they  are  parallel ;  hence  it  can  not  meet  the  plane  MN 
that  is,  AB  is  parallel  to  the  plane  MN  (Def.  2).     Therefore   ' 
if  a  straight  line  &c. 


BOOK    VII. 


119 


PROPOSITION  XI.       TFEOREM. 

Two  planes,  which  are  perpendicular  to  the  same  straight 
line,  are  parallel  to  each  other. 

Let  the  planes  MN,  PQ  be 
perpendicular  to  the  line  AB ; 
then  will  they  be  parallel  to  each 
other. 

For  if  they  are  not  parallel, 
they  will  meet  if  produced.  Let 
them  be  produced  and  meet  in  C. 
Join  AC,  BC.  Now  the  line  AB, 
which  is  perpendicular  to  the  plane  MN,  is  perpendicular  to 
the  line  AC  drawn  through  its  foot  in  that  plane.  For  the 
same  reason  AB  is  perpendicular  to  BC.  Therefore  CA  and 
CB  are  two  perpendiculars  let  fall  from  the  same  point  C 
upon  the  same  straight  line  AB,  which  is  impossible  (Prop. 
XVI.,  B.  I.).  Hence  the  planes  MN,  PQ,  can  not  meet  when 
produced ;  that  is,  they  are  parallel  to  each  other.  There- 
fore, two  planes,  &c. 


PROPOSITION    XII.       THEOREM. 

If  two  parallel  planes  are  cut  by  a  third  plane,  their  common 
sections  are  parallel. 

Let  the  parallel  planes  MN,  PQ  be 
cut  by  the  plane  ABDC ;  and  let  their 
common  sections  with  it  be  AB,  CD ; 
then  will  AB  be  parallel  to  CD. 

For  the  two  lines  AB,  CD  are  in  the 
same  plane,  viz.,  in  the  plane  ABDC 
which  cuts  the  planes  MN,  PQ;  and 
if  these  lines  were  not  parallel,  they 
would  meet  when  produced ;  therefore 

the  planes  MN,  PQ  would  also  meet,  which  is  mpossible,  be- 
cause they  are  parallel.  Hence  the  lines  AB.  CD  are  paral 
lei.  Therefore,  if  two  para  lei  planes,  &c. 


120 


GEOMETRY. 


PROPOSITION   XIII.       THEOREM. 

If  two  planes  are  parallel,  a  straight  line  which  is  perpen 
dicular  to  one  of  them,  is  also  perpendicular  to  the  other. 

Let  the  two  planes  MN,  PQ  be  par- 
allel, and  let  the  straight  line  AB  be 
perpendicular  to  the  plane  MN ;  AB 
will  also  be  perpendicular  to  the  plane 

PQ. 

Through  the  point  B,  draw  any  line 
BD  in  the  plane  PQ ;  and  through  the 
lines  AB,  BD  suppose  a  plane  to  pass  intersecting  the  plane- 
MN  in  AC.  The  two  lines  AC,  BD  will  be  parallel  (Prop. 
XII.).  But  the  line  AB,  being  perpendicular  to  the  plane 
MN,  is  perpendicular  to  the  straight  line  AC  which  it  meets 
in  that  plane  ;  it  must,  therefore,  be  perpendicular  to  its  par- 
allel BD  (Prop.  XXIIL,  Cor.  1,  B.  I.).  But  BD  is  any  line 
drawn  through  B  in  the  plane  PQ, ;  and  since  AB  is  perpen- 
dicular to  any  line  drawn  through  its  foot  in  the  plane  PQ, 
it  must  be  perpendicular  to  the  plane  PQ  (Def.  1).  There 
rore,  if  two  planes,  &c. 


PROPOSITION    XIV.       THEOREM. 

Parallel  straight  lines  included  between  two  parallel  plane* 
fire  equal. 

Let  AB,  CD  be  the  two  parallel 
straight  lines  included  between  two 
parallel  planes  MN,  PQ ;  then  will  AB 
be  equal  to  CD. 

Through  the  two  parallel  lines  AB, 
CD  suppose  a  plane  ABDC  to  pass,  in- 
tersecting the  parallel  planes  in  AC  and 
BD.  The  lines  AC,  BD  will  be  parallel  to  each  other  (Prop. 
XII.).  But  AB  is,  by  supposition,  parallel  to  CD  ;  therefore 
the  figure  ABDC  is  a  parallelogram ;  and,  consequently,  AB 
is  equal  to  CD  (Prop.  XXIX.,  B.  L).  Therefore,  parallel 
straight  lines,  &c. 

Cor.  Hence  two  parallel  planes  are  every  where  equidis- 
tant; for  if  AB,  CD  are  perpendicular  to  the  plane  MTV,  they 
will  be  perpendicular  to  the  parallel  plane  PQ  (Prop.  XIII.) ; 
and  being  both  perpendicular  to  the  same  plane,  they  will  be 
parallel  to  each  other  (Prop  IX.),  and,  consequently,  equal 


BOOK    VII. 


PROPOSITION    XV.       THEOREM. 


If  two  angles,  not  in  tk?  same  plane,  have  their  sides 
parallel  and  similarly  situated,  these  angles  will  be 
and  their  planes  will  be  parallel. 

Let  the  two  angles  ABC,  DEF,  lying 
in  different  planes  MN,  PQ,  have  their 
sides  parallel  each  to  each  and  similarly 
situated  ;  then  will  the  angle  ABC  be 
equal  to  the  angle  DEF,  and  the  plane 
MN  be  parallel  to  the  plane  PQ,. 

Take  AB  equal  to  DE,  and  BC  equal 
to  EF,  and  join  AD,  BE,  CF,  AC,  DF. 
Then,  because  AB  is  equal  and  parallel  to  DE,  the  figure 
ABED  is  a  parallelogram  (Prop.  XXXL,  BI.)  ;  and  AD  is 
equal  and  parallel  to  BE.  For  the  same  reason  CF  is  equal 
and  parallel  to  BE.  Consequently,  AD  and  CF,  being  each 
of  them  equal  and  parallel  to  BE,  are  parallel  to  each  other 
(Prop.  IX.,  Cor.  2),  and  also  equal  ;  therefore  AC  is  also  equal 
and  parallel  to  DF  (Prop.  XXXI.,  B.  I.).  Hence  the  triangles 
ABC,  DEF  are  mutually  equilateral,  and  the  angle  ABC  is 
equal  to  the  angle  DEF  (Prop.  XV.,  B.  I.). 

Also,  the  plane  ABC  is  parallel  to  the  plane  DEF.  For, 
if  they  are  not  parallel,  suppose  a  piano  to  pass  through  A 
parallel  to  DEF,  and  let  it  meet  the'  straight  lines  BE,  CF  in 
the  points  G-  and  H.  Then  the  three  lines  AD,  G-B,  HF  will 
be  equal  (Prop.  XIV.).  But  the  three  lines  AD,  BE,  CF  havo 
already  been  proved  to  be  equal;  hence  BE  is  equal  to  GrE, 
and  CF  is  equal  to  HF,  which  is  absurd  ;  consequently,  the 
plane  ABC  must  be  paralM  to  the  plane  DEF.  Therefore, 
if  two  angles,  &c. 

Cor.  1.  If  two  parallel  planes  MN,  PQ,  are  met  by  two 
other  planes  ABED,  BCFE,  the  angles  formed  by  the  inter- 
sections of  the  parallel  planes  will  be  equal.  For  the  section 
AB  is  parallel  to  the  section  DE  (Prop.  XII.)  ;  and  BC  is 
parallel  to  EF  ;  therefore,  by  the  Proposition,  the  angle  ABC 
is  equal  to  the  angle  DEF. 

Cor.  2.  If  three  straight  lines  AD,  BE,  CF,  not  situated  in 
the  same  plane,  are  equal  and  parallel,  the  triangles  ABC/ 
DEF,  formed  by  joining  the  extremities  of  these  lines,  will 
be  equal,  and  their  planes  will  be  parallel.  For,  since  AD  i3 
equal  and  parallel  to  BE,  the  figure  ABED  is  a  parallel- 
ogram ;  hence  the  side  AB  is  equal  and  parallel  to  DE,  For 


122 


GEOMETRY. 


the  same  reason,  the  sides  BC  and  EF  are  equal  and  parai 
lei;  as,  also,  the  sides  AC  and  DF.  Consequently,  the  two 
triangles  ABC,  DEF  are  equal ;  and,  according  to  the  Prop- 
osition, their  planes  are  parallel. 


PROPOSITION    XVI.       THEOREM. 

If  two  straight  lines  are  cut  by  parallel  planes,  they  will  b& 
cut  in  the  same  ratio. 

Let  the  straight  lines  AB,  CD  be  cut 
by  the  parallel  planes  MN,  PQ,  RS  in 
the  points  A,  E,  B,  C,  F,  D ;  then  we 
shall  have  the  proportion 

AE  :  EB  :  :  CF  :  FD. 
Draw  the  line  BC  meeting  the  plane 
PQ  in  G,  and  join  AC,  BD,  EG,  GF. 
Then,  because  the  two  parallel  planes 
MN,  PQ  are  cut  by  the  plane  ABC,  the 
common  sections  AC,  EG  are  parallel  (Prop.  XII.).    Also,  be 
cause  the  two  parallel  planes  PQ,  RS  are  cut  by  the  plane 
BCD,  the  common  sections  BD,  GF  are  parallel.     Now,  be- 
cause EG  is  parallel  to  AC,  a  side  of  the  triangle  ABC  (Prop. 
XVI.,  B.  IV.),  we  have 

AE  :  EB  :  :  CG  :  GB. 

Also,  because  GF  is  parallel  to  BD,  one  side  of  the  triangle 
BCD,  we  have 

CG  :  GB  :  :  CF  :  FD ; 
hence  (Prop.  IV.,  B.  II.), 

AE  :  EB  :  :  CF  :  FD. 
Therefore,  if  two  straight  lines,  &c. 


PROPOSITION   XVII.       THEOREM. 

If  a  solid  angle  is  contained  by  three  plane  angles,  tfte  fum 
of  any  two  of  these  angles  is  greater  than  the  third. 

Let  the  solid  angle  at  A  be  con- 
tained by  the  three  plane  angles 
BAG,  CAD,  DAB;  any  two  of  these 
angles  will  be  greater  than  the  third. 

If  these  three  angles  are  all  equal 
to  each  other,  it  is  plain  that  any 
two  of  them  must  be  greater  than 
the  third.  But  if  they  are  not  equal 


BOOK     Vii.  123 

let  BAG  be  that  angle  wnich  is  no  less  than  either  of  the 
other  two,  and  is  greater  than  one  of  them  BAD.  Then,  at 
the  point  A,  make  the  angle  BAE  equal  to  the  angle  BAD ; 
take  AE  equal  to  AD ;  through  E  draw  the  line  BEC  cutting 
AB,  AC  in  the  points  B  and  C  ;  and  join  DB,  DC. 

Now,  because,  in  the  two  triangles  BAD,  BAE,  AD  is 
equal  to  AE,  AB  is  common  to  both,  and  the  angle  BAD  is 
equal  to  the  angle  BAE ;  therefore  the  base  BD  is  equal  to 
the  base  BE  (Prop.  VI.,  B.  I.).  Also,  because  the  sum  of  the 
lines  BD,  DC  is  greater  than  BC  (Prop.  VIII.,  B.  L),  and  BD 
is  proved  equal  to  BE,  a  part  of  BC,  therefore  the  remaining 
line  DC  is  greater  than  EC.  Now,  in  the  two  triangles 
CAD,  CAE,  because  AD  is  equal  to  AE,  AC  is  common,  but 
the  base  CD  is  greater  than  the  base  CE ;  therefore  the  an 

e  CAD  is  greater  than  the  angle  CAE  (Prop.  XIV.,  B.  L). 

ut,  by  construction,  the  angle  BAD  is  equal  to  the  angle 
BAE ;  therefore  the  two  angles  BAD,  CAD  are  together 
greater  than  BAE,  CAE ;  that  is,  than  the  angle  BAG. 
Now  BAG  is  not  less  than  either  of  the  angles  BAD,  CAD  ; 
hence  BAG,  with  either  of  them,  is  greater  than  the  third. 
Therefore,  if  a  solid  angle,  &c. 


PROPOSITION    XVIII.       THEOREM. 

The  plane  angles  which  contain  any  solid  angle,  are  togethe* 
less  than  four  right  angles. 

Let  A  be  a  solid  angle  contained  by  any 
number  of  plane  angles  BAG,  CAD,  DAE, 
EAF,  FAB ;  these  angles  are  together  less 
than  four  right  angles. 

Let  the  planes  which  contain  the  solid  an- 
gle at  A  be  cut  by  another  plane,  forming 
the  polygon  BCDEF.  Now,  because  the 
solid  angle  at  B  is  contained  by  three  plane 
angles,  any  two  of  which  are  greater  than 
the  third  (Prop.  XVIL),  the  two  angles  ABC, 
ABF  are  greater  than  the  angle  FBC.  For 
the  same  reason,  the  two  angles  ACB,  ACD  are  greater  than 
the  angle  BCD,  and  so  with  the  other  angles  of  the  polygon 
BCDEF.  Hence,  the  sum  of  all  the  angles  at  the  bases  of  the 
triangles  having  the  common  vertex  A,  is  greater  than  the 
sum  of  all  the  angles  of  the  polygon  BCDEF.  But  all  the 
angles  of  these  triangles  are  together  equal  to  twice  as  many 
right  angles  as  there  are  triangles  (Prop.  XXVIL,  B.  L),  that 
*s.  ns  there  are  sides  of  the  polygon  BCDEF.  Also,  ths  an* 


124 


GEOMETRT. 


gles  of  the  polygon,  together  with  lour  right  angles,  are  equal 
to  twice  as  many  right  angles  as  the  figure  has  sides  (Prop. 
XXVIIL,  B.  I.) ;  hence  all  the  angles  of  the  triangles  are 
equal  to  all  the  angles  of  the  polygon,  together  with  four 
right  angles.  But  it  has  been  proved  that  the  angles  at  the 
oases  of  the  triangles,  are  greater  than  the  angles  of  the 
polygon.  Hence  the  remaining  angles  of  the  triangles,  viz., 
those  which  contain  the  solid  angle  at  A,  are  less  than  four 
right  angles.  Therefore,  the  plane  angles,  &c. 

Scholium.  This  demonstration  supposes  that  the  solid  an- 
gle is  convex ;  that  is,  that  the  plane  of  neither  of  the  faces, 
if  produced,  wouid  cut  the  solid  angle.  If  it  were  otherwise, 
the  sum  of  the  plane  angles  would  no  longer  be  limited,  and 
might  be  of  any  magnitude. 


PROPOSITION    XIX.       THEOREM. 

If  two  solid  angles  are  contained  by  three  plane  angles  which 
are  equal,  each  to  each,  the  planes  of  the  equal  angles  will  be 
equally  inclined  to  each  other. 

Let  A  and  a  be  two  solid 
angles,  contained  by  three 
plane  angles  which  are 
equal,  each  to  each,  viz.,  the 
angle  BAG  equal  to  bac, 
the  angle  CAD  to  cad,  and 
BAD  equal  to  bad;  then  B 
will  the  inclination  of  the 
planes  ABC,  ABD  be  equal 
to  the  inclination  of  the  planes  abc,  abd. 

In  the  line  AC,  the  common  section  of  the  planes  ABC, 
ACD,  take  any  point  C ;  and  through  C  let  a  plane  BCE 
pass  perpendicular  to  AB,  and  another  plane  CDE  perpen- 
dicular to  AD.  Also,  take  ac  equal  to  AC ;  and  through  c 
let  a  plane  bee  pass  perpendicular  to  ab,  and  another  plane 
rde  perpendicular  to  ad. 

Now,  since  the  line  AB  is  perpendicular  to  the  plane  BCE, 
it  is  perpendicular  to  every  straight  line  which  it  meets  in 
that  plane  ;  hence  ABC  and  ABE  are  right  angles.  For  the 
same  reason  abc  and  abe  are  right  angles.  Now,  in  the  tri 
angles  ABC,  abc,  the  angle  BAG  is,  by  hypothesis,  equal  to 
bac,  and  the  angles  ABC,  abc  are  right  angles ;  therefore  the 
angles  ACB,  acb  are  equal.  But  the  side  AC  was  made 
equal  to  the  side  ac;  hence  the  two  triangles  are  equal 
(JNop.  VII.,  U.  U  ;  that  is,  the  side  AB  is  equal  to  ab,  and  BC 


BOOK    VII. 

lo  be.     In  the  same  manner,  it  may  be  proved  that  AD  ia 
equal  to  ad,  and  CD  to  cd. 

We  can  now  prove  that  the  quadrilateral  ABED  is  equal 
to  the  quadrilateral  abed.  For,  let  the  angle  BAD  be  placed 
upon  the  equal  angle  bad,  then  the  point  B  will  fall  upon  the 
point  b,  and  the  point  D  upon  the  point  d;  because  AB  ia 
equal  to  ab,  and  AD  to  ad.  At  the  same  time.  BE,  which  ia 
perpendicular  to  AB,  will  fall  upon  be,  which  is  perpendicu 
Jar  to  ab ;  and  for  a  similar  reason  DE  will  fall  upon  de 
Hence  the  point  E  will  fall  upon  e,  and  we  shall  have  BE 
equal  to  be,  and  DE  equal  to  de. 

Now,  since  the  plane  BCE  is  perpendicular  to  the  line  AB, 
it  is  perpendicular  to  the  plane  ABD  which  passes  through 
AB  (Prop.  VI.) .  For  the  same  reason  CDE  is  perpendicular 
to  the  same  plane ;  hence  CE,  their  common  section,  is  per- 
pendicular to  the  plane  ABD  (Prop.  VIII.).  In  the  same  man- 
ner, it  may  be  proved  that  ce  is  perpendicular  to  the  plane 
abd.  Now,  in  the  triangles  BCE,  6ce,the  angles  BEC,  bee  are 
right  angles,  the  hypothenuse  BC  is  equal  to  the  hypothenuse 
be,  and  the  side  BE  is  equal  to  be  ;  hence  the  two  triangles 
are  equal,  and  the  angle  CBE  is  equal  to  the  angle  cbe.  But 
the  angle  CBE  is  the  inclination  of  the  planes  ABC,  ABD 
(Def.  4)  ;  and  the  angle  cbe  is  the  inclination  of  the  planes 
abc,  abd;  hence  these  planes  are  equally  inclined  to  each 
other. 

We  must,  however,  observe  that  the  angle  CBE  is  not, 
propei'ly  speaking,  the  inclination  of  the  planes  ABC,  ABD, 
except  when  the  perpendicular  CE  falls  upon  the  same  side 
of  AB  is  AD  does.  If  it  fall  upon  the  other  side  of  AB,  then 
the  angle  between  the  two  planes  will  be  obtuse,  and  this 
angle,  together  with  the  angle  B  of  the  triangle  CBE,  will 
make  two  right  angles.  But  in  this  case,  the  angle  between 
the  two  planes  abc,  abd  will  also  be  obtuse,  and  this  angle, 
together  with  the  angle  b  of  the  triangle  cbe,  will  also  make 
two  right  angles.  And,  since  the  angle  B  is  always  equal  to 
the  angle  b,  the  inclination  of  the  two  planes  ABC,  ABD  will 
always  be  equal  to  that  of  the  planes  abc,  abd.  Therefore,  if 
;wo  solid  angles,  &c. 

Scholium.  If  two  solid  angles  are  contained  by  three 
plane  angles  which  are  equal,  each  to  each,  and  similarly 
situated,  the  angles  will  be  equal,  and  will  coincide  when 
applied  the  one  to  the  other.  For  we  have  proved  that  the 
quadrilateral  ABED  will  coincide  with  its  equal  abed 
Now,  because  the  triangle  BCE  is  equal  to  the  triangle  bee, 
the  line  CE,  which  is  perpendicular  to  the  plane  ABED,  ia 
equal  to  the  line  ce,  which  is  perpendicular  to  the  plane  abed. 
And  since  only  one  perpendicular  can  be  drawn  to  a  plane 


126 


GEOMETRY. 


From  the  same  point  (Prop. 
IV.,  Cor.  2),  the  lines  CE,  ce 
must  coincide  with  each  oth- 
er, and  the  point  C  coincide 
with  the  point  c.  Hence 
the  two  solid  angles  must 
coincide  throughout. 

It  should,  however,  be  ob- 
served that  the  two  solid  an- 
gles do  not  admit  of  superposition,  unless  the  three  equal  plane 
angles  are  similarly  situated  in  both  cases.  For  if  the  per- 
pendiculars CE,  ce  lay  on  opposite  sides  of  the  planes  ABED 
abed,  the  two  solid  angles  could  not  be  made  to  coincide 
Nevertheless,  the  Proposition  will  always  hold  true,  that  the 
planes  containing  the  equal  angles  are  equally  inclined  to 
other. 


BOOK   VIII.  12' 


BOOK  VIII. 

POLYEDRONS 

Definitions. 

1 .  A  polyedron  is  a  solid  included  by  any  number  of  planes 
which  are  called  its  faces.     If  the  solid  have  only  four  faces, 
which  is  the  least  number  possible,  it  is  called  a  tetraedron , 
if  six  faces,  it  is  called  a  hexaedron  ;  if  eight,  an  octaedron  • 
if  twelve,  a  dodecaedron  ;  if  twenty,  an  icosaedron,  &c. 

2.  The  intersections  of  the  faces  of  a  polyedron  are  called 
*ts  edges.     A  diagonal  of  a  polyedron  is  the  straight  line 
which  joins  any  two  vertices  not  lying  in  the  same  face. 

3.  Similar  polyedrons  are  such  as  have  all  their  solid  an- 
gles equal,  each  to  each,  and  are  contained  by  the  same  num- 
ber of  similar  polygons. 

4.  A  regular  polyedron  is  one  whose  solid  angles  are  all 
equal  to  each  other,  and  whose  faces  are  all  equal  and  regu 
lar  polygons. 

5.  A  prism  is  a  polyedron  having  two  faces 
which  are  equal  and  parallel  polygons  ;  and 
the  others   are   parallelograms.     The   equal 
and  parallel  polygons  are  called  the  bases  of 
the  prism ;   the   other  faces   taken  together 
form  the  lateral  or  convex  surface.     The  alti- 
tude of  a  prism  is  the  perpendicular  distance 
between  its  two  bases.     The  edges  which  join 
the  corresponding  angles  of  the  two  polygons 
are  called  the  principal  edges  of  the  prism. 

6.  A  right  prism  is  one  whose  principal  edges  are  all  pei 
pendicular  to  the  bases.     Any  other  prism  is  called  an  ob- 
lique prism 

7.  A  prism  is  triangular,  quadrangular,  pentagonal,  hex- 
agonal, &c.,  according  as  its  base  is  a  triangle,  a  quadri- 
lateral, a  pentagon,  a  hexagon,  &c. 

8.  A  parallelepiped  is  a  prism  whose 
bases  are  parallelograms.     A  right  par- 

allelopiped  is  one  whose  faces  are  all  rect-     M  .™/ 

angles.  /--""          L^ 

9.  A  cube,  is  a  nght  parallelepiped  bounded  by  six  equac 
squares. 


128 


GEOMETRY 


10.  A  pyramid  is  a  polyedron  contained  by 
several  triangular  planes  proceeding  from  the 
same  point,  and  terminating  in  the  sides  of  a 
polygon.     This  polygon  is  called  the  base  of 
the  pyramid  ;  and  the  point  in  which  the  planes 
meet,  is  the  vertex.    The  triangular  planes  form 
the  convex  surface. 

11.  The  altitude  of  a  pyramid  is  the  perpen- 
dicular let  fall  from  the  vertex  upon  the  plane 

of  the  base,  produced  if  necessary.  The  slant  height  of  a 
pyramid  is  a  line  drawn  from  the  vertex,  perpendicular  to 
one  side  of  the  polygon  which  forms  its  base. 

12.  A  pyramid  is  triangular  ',  quadrangular,  &c.,  according 
as  the  base  is  a  triangle,  a  quadrilateral,  &c. 

13.  A  regular  pyramid  is  one  whose  base  is  a  regular  poly- 
gon, and  the  perpendicular  let  fall  from  the  vertex  upon  the 
base,  passes  through  the  center  of  the  base.     This  perpendic- 
ular is  called  the  axis  of  the  pyramid. 

14.  A  frustum  of  a  pyramid  is  a  portion  of  the  solid  next 
the  base,  cut  off  by  a  plane  parallel  to  the  base.     The  alti- 
tude of  the  frustum    is  the  perpendicular  distance  between 
the  two  parallel  planes. 

PROPOSITION    I.       THEOREM. 

The  convex  surface  of  a  right  prism  is  equal  to  the  pe* 
rimeter  of  its  base  multiplied  by  its  altitude. 

Let  ABCDE-K  be  a  right  prism  ;  then  will 
its  convex  surface  be  equal  to  the  perimeter 
of  the  base  of  AB+BC-f  CD+DE  +  EA  multi-  E| 
plied  by  its  altitude  AF. 

For  the  convex  surface  of  the  prism  is 
equal  to  the  sum  of  the  parallelograms  AGr, 
BH,  CI,  &c.  Now  the  area  of  the  parallelo- 
gram  AGr  is  measured  by  the  product  of  its 
base  AB  by  its  altitude  AF  (Prop.  IV.,  Sch., 
B.  IV.).  The  area  of  the  parallelogram  BH  is  measured  by 
BCxBGr;  the  area  of  CI  is  measured  by  CDxCH,  and  so 
of  the  others.  But  the  lines  AF,  BG-,  CH,  &c.,  are  all  equal 
to  each  other  (Prop.  XIV.,  B.  VII.),  and  each  equal  to  the 
altitude  of  the  prism.  Also,  the  lines  AB,  BC,  CD,  &c.,  taken 
together,  from  the  perimeter  of  the  base  of  the  prism.  There- 
fore, the  sum  of  these  parallelograms,  or  the  convex  surface 
of  the  prism,  is  equal  to  the  perimeter  of  its  base,  multiplied 
by  its  altitude. 


BOOK    VIII. 


Cat'.  lf  two  right  prisms  have  the  same  altitude,  theii 
convex  surfaces  will  be  to  each  other  as  the  perimeters  of 
their  bases. 


PROPOSITION    II.       THEOREM. 

In  every  prism,  the  sections  formed  by  parallel  planes  art 
equal  polygons. 

Let  the  prism  LP  be  cut  by  the  parallel 
planes  AC,  FH  ;  then  will  the  sections  ABC 
DE,  FG-HIK,  be  equal  polygons. 

Since  AB  and  FGr  are  the  intersections 
of  two  parallel  planes,  with  a  third  plane 
LMON,  they  are  parallel.  The  lines  AF, 
BG-  are  also  parallel,  being  edges  of  the 
prism;  therefore  ABGrF  is  a  parallelogram,-.-! 
and  AB  is  equal  to  FGr.  For  the  same 
reason  BC  is  equal  and  parallel  to  GrH,  CD 
to  IH,  DE  to  IK,  and  AE  to  FK. 

Because  the  sides  of  the  angle  ABC  are  parallel  to  those  of 
FG-H,  and  are  similarly  situated,  the  angle  ABC  is  equal  to 
FGrH  (Prop.  XV.,  B.  VII.).  In  like  manner  it  may  be  proved 
that  the  angle  BCD  is  equal  to  the  angle  GrHI,  and  so  of  the 
rest.  Therefore  the  polygons  ABCDE,  FGrHIK  are  equal. 

Cor.  Every  section  of  a  prism,  made  parallel  to  the  base, 
is  equal  to  the  base. 


M 


PROPOSITION    III.       THEOREM. 

Two  prisms  are  equal,  when  they  have  a  solid  angle  con- 
tained by  three  faces  which  are  equal,  each  to  each,  ana 
similarly  situated. 

Let  AI,  ai  be  two  prisms 
having  the  faces  which  con- 
tain the  solid  angle  B  equal 
to  the  faces  which  contain 
the  solid  angle  b ;  viz.,  the 
base  ABCDE  to  the  base 
abode,  the  parallelogram 
AG-  to  the  parallelogram 
ag,  and  the  parallelogram 
BH  to  the  parallelogram  bh 
to  the  prism  ai. 


B        c 
then  will  the  prism  AI  be  equaJ 


j  30  GEOMETRY, 

Let  ihe  prism  AI  be 
applied  to  t\ie  prism  ai,  so 
that  the  equal  bases  AD 
and  ad  may  coincide,  the 
point  A  falling  upon  a,  B 
upon  b,  and  so  on.  And 
because  the  three  plane 
angles  which  contain  the 
solid  angle  B,  are  equal 
to  the  three  plane  angles 
which  contain  the  solid  angle  b,  and  these  planes  are  similarly 
situated,  the  solid  angles  B  and  b  are  equal  (Prop.  XIX.,  Sch. 
B.  VII.).  Hence  the  edge  BG  will  coincide  with  its  equal  bg  • 
and  the  point  G  will  coincide  with  the  point  g.  Now,  be- 
cause the  parallelograms  AG  and  ag  are  equal,  the  side  Gt 
will  fall  upon  its  equal  gf;  and  for  the  same  reason,  GH  wiL 
fall  upon  gh.  Hence  the  plane  of  the  base  FGHIK  will  coin- 
cide with  the  plane  of  the  basefghik  (Prop.  II.,  B.  VII.).  But 
since  the  upper  bases  are  equal  to  their  corresponding  lowei 
bases,  they  are  equal  to  each  other  ;  therefore  the  base  FI  will 
coincide  throughout  with/i;  viz.,  HI  with  hi,  IK  with  ik,  and 
KF  with  kf;  hence  the  prisms  coincide  throughout,  and  are 
equal  to  each  other.  Therefore,  two  prisms,  &c. 

Cor.  Two  right  prisms,  which  have  equal  bases  and  equal 
altitudes,  are  equal. 

For,  since  the  side  AB  is  equal  to  ab,  and  the  altitude  BG 
to  bg,  the  rectangle  ABGF  is  equal  to  the  rectangle  abgf, 
So,  also,  the  rectangle  BGHC  is  equal  to  the  rectangle  bghc  ; 
hence  the  three  faces  which  contain  the  solid  angle  B  are 
equal  to  the  three  faces  which  contain  the  solid  angle  b  • 
consequently,  the  two  prisms  are  equal. 


PROPOSITION  IV.       THEOREM. 

The  opposite  faces  of  a  parallelepiped  are  equal  and  parallel 

Let  ABGH  be  a  parallelepiped ;  then 
will  its  opposite  faces  be  equal  and  parallel. 

From  the  definition  of  a  parallelopiped 
(Def.  8)  the  bases  AC,  EG  are  equal  and 
parallel ;  and  it  remains  to  be  proved  that 
the  same  is  true  of  any  two  opposite  faces, 

as  AH,  BG.     Now,  because  AC  is  a  par-        __ . 

allelogram,  the  side  AD  is  equal  and  par- 
allel to  BC.     For  the  same  reason  AE  is  equal  and  parallel 
to  BF ;  hence    he  angle  DAE  is  equal  to  the  angle  CBF 


UUOK     VIII. 


131 


(Prop,  XV.,  B.  VII.),  and  the  plane  DAE  is  parallel  to  the 
plane  CBF.  Therefore  also  the  parallelogram  AH  is  equal 
to  the  parallelogram  BG.  In  the  same  manner,  it  may  be 
proved  that  the  opposite  faces  AF  and  DG  are  equal  and 
parallel.  Therefore,  the  opposite  faces,  &c. 

Cor.  1.  Since  a  parallelepiped  is  a  solid  contained  by  six 
faces,  of  which  the  opposite  ones  are  equal  and  parallel,  any 
face  may  be  assumed  as  the  base  of  a  parallelepiped. 

Cor.  2.  The  four  diagonals  of  a  parallelepiped  bisect  each 
other. 

Draw  any  two  diagonals  AG,  EC ;  they 
will  bisect  each  other.  Since  AE  is  equal 
and  parallel  to  CG,  the  figure  AEGC  is  a 
parallelogram ;  and  therefore  the  diago- 
nals AG,  EC  bisect  each  other  (Prop. 
XXXIL,  B.  L).  In  the  same  manner,  it 
may  be  proved  that  the  two  diagonals  BH 
and  DF  bisect  each  other;  and  hence  the 
four  diagonals  mutually  bisect  each  other,  in  a  point  whicn 
may  be  regarded  as  the  center  of  the  parallelepiped. 


PROPOSITION   V.       THEOREM. 


If  a  parallelopiped  be  cut  by  a  plane  passing  through  the 
diagonals  of  two  opposite  faces,  it  will  be  divided  into  two 
equivalent  prisms. 


Let  AG  be  a  parallelopiped,  and  AC, 
EG  the  diagonals  of  the  opposite  parallelo- 
grams  BD,  FH.  Now,  because  AE,  CG  are 
each  of  them  parallel  to  BF,  they  are  par- 
allel  to  each  other  ;  therefore  the  diagonals 
AC,  EG  are  in  the  same  plane  with  AE, 
CG  ;  and  the  plane  AEGC  divides  the  solid 
AG  into  two  equivalent  prisms. 

Through  the  vertices  A  and  E  draw  the 
planes  AIKL,  EMNO  perpendicular  to  AE, 
meeting  the  other  edges  of  the  parallelo- 
piped in  the  points  I,  K,  L,  and  in  M,  N,  O. 
The  sections  AIKL,  EMNO  are  equal,  because  they  are 
formed  by  planes  perpendicular  to  the  same  straight  line, 
and,  consequently,  parallel  (Prop.  II.).  They  are  also  par- 
allelograms, because  AI,  KL,  two  opposite  sides  of  the  same 
section,  are  the  intersections  of  two  parallel  planes  ABFE, 
DCGH,by  the  same  plane. 

For  the  same  reason,  the  figure  ALOE  is  a  parallelogram  ; 


132  GEOMETRY. 

so,  also,   are  AIME,   IKNM,    KLON,   the 
other   lateral    faces   of    the    solid   AIKL-H 
EMNO  ;  hence  this  solid  is  a  prism  (Def. 
5) ;  and  it  is  a  right  prism  because  AE  is 
perpendicular  to  the  plane  of  its  base.     But 
the  right  prism  AN  is   divided   into  two 
equal    prisms   ALK-N,   AIK-N ;    for   theD 
basis  of  these  prisms  are  equal,  being  halves  L'-^ 
cf  the  same  parallelogram  AIKL,  and  they 
have  the  common  altitude  AE  ;    they  are 
therefore  equal  (Prop.  III.  Cor.). 

Now,  because  AEHD,  AEOL  are  parallelograms,  the  sided 
DH,  LO,  being  equal  to  AE,  are  equal  to  each  other.  Take 
away  the  common  part  DO,  and  we  have  DL  equal  to  HO. 
For  the  same  reason,  CK  is  equal  to  G-N.  Conceive  now  that 
ENO,  the  base  of  the  solid  ENG-HO,  is  placed  on  AKL,  the 
base  of  the  solid  AKCDL  ;  then  the  point  0  falling  on  L  and 
N  on  K,  the  lines  HO,  GN  will  coincide  with  their  equals 
DL,  CK,  because  they  are  perpendiculars  to  the  same  plane 
Hence  the  two  solids  coincide  throughout,  and  are  equal  to 
each  other.  To  each  of  these  equals,  add  the  solid  ADC-N ; 
then  will  the  oblique  prism  ADC-G-  be  equivalent  to  the 
right  prism  ALK-N. 

In  the  same  manner,  it  may  be  proved  that  the  oblique 
prism  ABC-G  is  equivalent  to  the  right  prism  AIK-N.  But 
the  two  right  prisms  have  been  proved  to  be  equal ;  hence 
the  two  oblique  prisms  ADC-G,  ABC-G  are  equivalent  to 
each  other.  Therefore,  if  a  parallelepiped,  &c. 

Cor.  Every  triangular  prism  is  half  of  a  parallelepiped 
having  the  same  solid  angle,  and  the  same  edges  AB,  BC,  BF. 

Scholium.  The  triangular  prisms  into  which  the  oblique 
parallelepiped  is  divided,  can  not  be  made  to  coincide,  because 
the  plane  angles  about  the  corresponding  solid  angles  are  not 
similarly  situated. 

PROPOSITION    VI.       THEOREM. 

Parallelopipeds,  of  the  same  base  and  the  same  altitude, 
Are  equivalent. 

Case  first.  When  their  upper  bases  are  between  the  samp 
parallel  lines. 

Let  the  parallelepipeds  AG-,  AL  haVe  the  base  AC  common, 
and  let  their  opposite  bases  EG,  IL  be  in  the  same  plane, 
and  between  the  same  parallels  EK,  HL  ;  then  will  the  solid 
AG-  be  equivalent  to  the  solid  AL. 


BOOK     VIII. 


133 


Because  AF,  AK  are  parallel- 
ograms, EF  and  IK  are  each 
equal  to  AB,  and  therefore  equal 
to  each  other.  Hence,  if  EF  and 
IK  be  taken  away  from  the  same 
line  EK,  the  remainders  El  and 
FK  will  be  equal.  Therefore 
the  triangle  AE1  is  equal  to  the 
triangle  BFK.  Also,  the  parallelogram  EM  is  equal  to  the 
parallelogram  FL,  and  AH  to  BG.  Hence  the  solid  angles 
at  E  and  F  are  contained  by  three  faces  which  are  equal  to 
each  other  and  similarly  situated  ;  therefore  the  prism  AEI- 
M  is  equal  to  the  prism  BFK-L  (Prop.  III.). 

Now,  if  from   the  whole   solid  AL,  we  take  the  prism 
AEI-M,  there  will  remain  the  parallelepiped  AL ;   and  if 
from  the  same  solid  AL,  we  take  the  prism  BFK-L,  there 
will  remain  the  parallelepiped  AG.     Hence  the  parallelopi 
peds  AL,  AG  are  equivalent  to  one  another. 

Case  second.  When  their  upper  bases  are  not  between  the 
same  parallel  lines. 

Let    the   parallelepipeds    AG,  op  M          T 

AL  have  the  same  base  AC  and  v - ^ 

the  same  altitude  ;  then  will  their 
opposite  bases  EG,  IL  be  in  the 
same  plane.  And,  since  the  sides 
EF  and  IK  are  equal  and  parallel 
to  AB,  they  are  equal  and  paral- 
lel to  each  other.  For  the  same 
reason  FG  is  equal  and  parallel 
to  KL.  Produce  the  sides  EH, 
FG,  as  also  IK,  LM,  and  let  A  B 

them  meet  in  the  points  N,  O,  P,  Q ;  the  figure  NOPQ  is  a 
parallelogram  equal  to  each  of  the  bases  EG,  IL ;  and,  con- 
sequently, equal  to  ABCD,  and  parallel  to  it. 

Conceive  now  a  third  parallelepiped  AP,  having  AC  for  its 
^wer  baseband  NP  for  its  upper  base.  The  solid  AP  will 
be  equivalent  to  the  solid  AG,  by  the  first  Case,  because  they 
have  the  same  lower  base,  and  their  upper  bases  are  in  the 
same  plane  and  between  the  same  parallels,  EQ,  FP.  For 
the  same  reason,  the  solid  AP  is  equivalent  to  the  solid  AL ; 
hence  the  solid  AG  is  equivalen.  to  ,he  sol;d  AL.  There- 
fore, parallelepipeds,  &c, 


134 


GEOMETRY, 


II 


PROPOSIT   ON    VII.       THEOREM. 

Any  paraU&opiped  is  equivalent  to  a  right  parallelepiped 
having  the  same  altitude  and  an  equivalent  base. 

Let  AL  be  any  parallelepiped ;  it  is  equivalent  to  a  right 
parallelepiped  having  the  same  altitude  and  an  equivalent 
base. 

From  the  points  A,  B,  C,  D  draw  AE,  BF,  CG,  DH,  per- 
pendicular to  the  plane  of  the  low- 
er base,  meeting  the  plane  of  the 
upper  base  in  the  points  E,  F,  G, 
H.  Join  EF,  FG,  GH,  HE  ;  there 
will  thus  be  formed  the  parallele- 
piped AG,  equivalent  to  AL  (Prop. 
VI.) ;  and  its  lateral  faces  AF,  BG, 
CH,  DE  are  rectangles.  If  the 
base  ABCD  is  also  a  rectangle, 
AG  will  be  a  right  parallelepiped, 
and  it  is  equivalent  to  the  parallel- 
epiped AL.  But  if  ABCD  is  not  a  rectangle,  from  A  and  B 
draw  AI,  BK  perpendicular  to  CD;  and  K]yr 
from  E  and  F  draw  EM,  FL  perpendicu- 
lar to  GH ;  and  join  IM,  KL.  The  solid 
ABKI-M  will  be  a  right  parallelepiped. 
For,  by  construction,  the  bases  ABKI  and 
EFLM  are  rectangles ;  so,  also,  are  the 
lateral  faces,  because  the  edges  AE,  BF. 
KL,  IM  are  perpendicular  to  the  plane  of 
the  base.  Therefore  the  solid  AL  is  a  right 
parallelepiped.  But  the  two  parallelepipeds 
AG,  AL  may  be  regarded  as  having  the  same  base  AF,  and 
the  same  altitude  AI ;  they  are  therefore  equivalent.  But  the 
parallelepiped  AG  is  equivalent  to  the  first  supposed  parallel- 
epiped ;  hence  this  parallelepiped  is  equivalent  to  the  ricfh 
para  lelopiped  AL,  having  the  same  altitude,  and  an  equiva 
lent  \  ase.  Therefore,  any  parallelepiped,  fcc. 


BOOK    VIII.  13ft 


PROPOSITION  VIII.       THEOREM. 

Right  paralhlopipeds,  having  the  same  base,  are  to  each  oth- 
er as  their  altitudes. 

Let  AG,  AL  be  two  right  parallelepipeds 
having  the  same  base  ABCD  ;  then  will  they 
be  to  each  other  as  their  altitudes  AE,  AL 

Case  first.  When  the  altitudes  are  in  the 
ratio  of  two  whole  numbers. 

Suppose  the  altitudes  AE,  AI  are  in  the 
latio  of  two  whole  numbers  ;  for  example,  as 
seven  to  four.  Divide  AE  into  seven  equal 
parts ;  AI  will  contain  four  of  those  parts. 
Through  the  several  points  of  division,  let 
planes  be  drawn  parallel  to  the  base ;  these 
planes  will  divide  the  solid  AG  into  seven  A 
small  parallelepipeds,  all  equal  to  each  other,  having  equal 
bases  and  equal  altitudes.  The  bases  are  equal,  because  ev- 
ery section  of  a  prism  parallel  to  the  base  is  equal  to  the  base 
(Prop.  II.,  Cor.) ;  the  altitudes  are  equal,  for  these  altitudes 
are  the  equal  divisions  of  the  edge  AE.  But  of  these  seven 
equal  parallelepipeds,  AL  contains  four ;  hence  the  solid  AG 
is  to  the  solid  AL,  as  seven  to  four,  or  as  the  altitude  AE  is 
to  the  altitude  AI. 

Case  second.  When  the  altitudes  are  not  in  the  ratio  of  two 
whole  numbers. 

Let  AG,  AL  be  two  parallelepipeds  whose  altitudes  have 
any  ratio  whatever ;  we  shall  still  have  the  proportion 
Solid  AG  :  solid  AL  :  :  AE  :  AI. 

For  if  this  proportion  is  not  true,  the  first  three  terms  re- 
maining the  same,  the  fourth  term  must  be  greater  or  less 
than  AI.     Suppose  it  to  be  greater,  and  that  we  have 
Solid  AG  :  solid  AL  :  :  AE  :  AO. 

Divide  AE  into  equal  parts  each  less  than  OI ;  there  will 
be  at  least  one  point  of  division  between  O  and  I.  Designate 
that  point  by  N.  Suppose  a  parallelepiped  to  be  construct- 
ed, having  ABCD  for  its  base,  and  AN  for  its  altitude  ;  and 
represent  this  parallelepiped  by  P.  Then,  because  the  alti- 
tudes AE,  AN  are  in  the  ratio  of  two  whole  numbers,  w« 
shall  have,  by  the  preceding  Case, 

Solid  AG  :  P  :  :  AE  :  AN. 
But,  by  hypothesis,  we  have 

Solid  AG  :  solid  AL  :  :  AE  :  AO. 
Hence  (Prop  IV.,  nor.,  B.  TI.)f 


QfioMETKY. 

Solid  AL  if:  :AO:AN. 

But  AO  is  greater  than  AN  ;  hence  the  solid  AL  must  be 
greater  than  P  (Def.  2,  B.  II.) ;  on  the  contrary,  it  is  lesss 
which  is  absurd.  Therefore  the  solid  AG-  can  not  be  to  the 
solid  AL,  as  the  line  AE  to  a  line  greater  than  AI. 

In  the  same  manner,  it  may  be  proved  that  the  fourth  term 
of  the  proportion  can  not  be  less  than  AI ;  hence  it  must  he 
AI,  and  we  have  the  proportion. 

Solid  AG- :  solid  AL  :  :  AE  :  AL 
Therefore,  right  parallelepipeds,  &c. 


PROPOSITION    IX.       THEOREM. 

Right  parallelepipeds,  having'  the  same  altitude,  are  to 
each  other  as  their  bases. 

Let  AG-,  AN  be  two  right  parallelepipeds  having  the  sam« 
altitude  AE  ;  then  will  they  be  to  each  other  as  their  bases; 
that  is, 

Solid  AG- :  solid  AN  :  :  base  ABCD  :  base  AIKL. 

Place  the  two  solids  so  that  their  M  E 

surfaces  may  have  the  common 
angle  BAE  ;  produce  the  plane 
LKNO  till  it  meets  the  plane  DCG-H 
in  the  line  PQ;  a  third  parallelopiped 
AQ  will  thus  be  formed,  which  may 
oe  compared  with  each  of  the  paral- 
lelopipeds  AG-,  AN.  The  two  solids 
AG-,  AQ,,  having  the  same  base 
AEHD,  are  to  each  other  as  their 
altitudes  AB,  AL  (Prop.  VIII.) ;  and 
the  two  solids  AQ,  AN,  having  the  same  base  ALOE,  are  to 
each  other  as  their  altitudes  AD,  AL  Hence  we  have  th* 
two  proportions 

Solid  AG  :  solid  AQ  :  :  AB  :  AL  ; 
Solid  AQ  :  solid  AN  :  :  AD  :  AI. 
Hence  (Prop.  XL,  Cor.,  B.  II.), 

Solid  Ad  :  solid  AN  :  :  ABxAD  :  ALx  AI. 
But  ABX  AD  is  the  measure  of  the  base  ABCD  (Prop.  IV., 
Sch.,  B.  IV.) ;  and  AL  X  AI  is  the  measure  of  the  base  AIKL  f 
hence 

Solid  AO  :  solid  AN  :  :  base  ABCD  :  base  AIKL 
Therefore,  right  parallelepipeds,  &c. 


BOOK    VIII.  137 


PROPOSITION    X        THEOREM;    - 

Any  two  right  parallelepipeds  are  to  each,  oilier  as  the  pro& 
ucts  of  their  bases  by  their  altitudes. 

Let  AG,  AQ  De  two  right  paral- 
lelopipeds,  of  which  the  bases  are 
the  rectangles  ABCD,  AIKL,  and 
(he  altitudes,  the  perpendiculars  AE, 
AP;  then  will  the  solid  AG  be  to 
the  solid  AQ,  as  the  product  of 
ABCD  by  AE,  is  to  the  product  of 
AIKL  by  AP. 

Place  the  two  solids  so  that  their      V-'"-^-^; 
surfaces  may  have  the  common  an-       3£          IjNl  J 
gle  BAE ;  produce  the  planes  ne- 
cessary to  form  the  third  parallele- 
piped AN,  having  the  same  base  with  AQ,  and  the  same  alti- 
tude with  AG.     Then,  by  the  last  Proposition,  we  shall  have 

Solid  AG  :  solid  AN  :  :  ABCD  :  AIKL. 
But  the  two  parallelepipeds  AN,  AQ,  having  the  same  base 
AIKL,  are  to  each  other  as  their  altitudes  AE,  AP  (Prop. 
VIII.)  ;  hence  we  have 

Solid  AN  :  solid  AQ  :  :  AE  :  AP. 

Comparing  these  two  proportions  (Prop.  ^L,  Cor.,  B.  II.) 
we  have 

Solid  AG  :  solid  AQ  :  :  ABCD  X  AE  :  AIKL  X  AP. 

If  instead  of  the  base  ABCD, we  put  its  equal  ABxAD, 
and  instead  of  AIKL,  we  put  its  equal  AI  X  AL,  we  shall  have 

Solid  AG  :  solid  AQ  :  :  ABx  ADx  AE  :  AIxALxAP. 
Therefore,  any  two  right  parallelepipeds,  &c. 

Scholium.  Hence  a  right  parallelepiped  is  measured  oy 
the  product  of  its  base  and  altitude,  or  the  product  of  its  iAree 
dimensions. 

It  should  be  remembered,  that  by  the  product  of  two  01 
more  lines,  we  understand  the  product  of  the  numbers  which 
represent  those  lines ;  and  these  numbers  depend  upon  the 
linear  unit  employed,  which  may  be  assumed  at  pleasure. 
If  we  take  a  foot  as  the  unit  of  measure,  then  the  number  of 
feet  in  the  length  of  the  base,  multiplied  by  the  number  of 
feet  in  its  breadth,  will  give  the  number  of  square  feet  in  the 
base.  If  we  multiply  this  product  by  the  number  of  feet  in 
the  altitude,  it  will  give  the  number  of  cubic  feet  in  the  par- 
allelopiped.  If  we  take  an  inch  as  the  unit  of  measure,  we 
shall  obtain  in  the  same  manner  the  number  of  cubic  inches 
in  the  parallelopiped. 


1  «S8  GEOMETRY. 


PROPOSITION   XI.       THEOREM. 

4 

The  solidity  of  a  prism  is  measured  by  the  product  of  it* 
base  by  its  altitude. 

For  any  parallelepiped  is  equivalent  to  a  right  parallele- 
piped, having  the  same  altitude  and  an  equivalent  base  (Prop. 
VII.).  But  the  solidity  of  the  latter,  is  measured  by  the  prod- 
uct of  its  base  by  its  altitude ;  therefore  the  solidity  of  the 
former  is  also  measured  by  the  product  of  its  base  by  its  al- 
titude. 

Now  a  triangular  prism  is  half  of  a  parallelepiped  having 
the  same  altitude  and  a  double  base  (Prop.  V.).  But  the 
solidity  of  the  latter  is  measured  by  the  product  of  its  base  by 
ts  altitude ;  hence  a  triangular  prism  is  measured  by  the 
product  of  its  base  by  its  altitude. 

But  any  prism  can  be  divided  into  as  many  triangular 
prisms  of  the  same  altitude,  as  there  are  triangles  in  the  poly- 
gon which  forms  its  base.  Also,  the  solidity  of  each  of  these 
triangular  prisms,  is  measured  by  the  product  of  its  base  by 
its  altitude ;  and  since  they  all  have  the  same  altitude,  the 
sum  of  these  prisms  will  be  measured  by  the  sum  of  the  tri- 
angles which  form  the  bases,  multiplied  by  the  common  alti- 
tude. Therefore,  the  solidity  of  any  prism  is  measured  by 
the  product  of  its  base  by  its  altitude. 

Cor.  If  two  prisms  have  the  same  altitude,  the  products  of 
ihe  bases  by  the  altitudes, will  be  as  the  bases  (Prop.  VIII., 
B.  II.)  ;  hence  prisms  of  the  same  altitude  are  to  each  other  as 
their  bases.  For  the  same  reason,  prisms  of  the  same  base 
are  to  each  other  as  their  altitudes  ;  and  prisms  generally  art 
to  each  other  as  the  products  of  their  bases  and  altitudes. 


PROPOSITION    XII.       THEOREM. 

Similar  prisms  are  to  each  other  as  the  cubes  of  their  homol 
vgous  edges. 

Let  ABCDE-F,  abcde-f  be  two  similar  prisms ;  then  wil. 
the  prism  AD-F  be  to  the  prism  ad-f,  as  AB3  to  ab*,  or  as 
AFS  to  af. 

For  the  solids  are  to  each  other  as  the  products  of  their 
bases  and  altitudes  (Prop.  XL,  Cor.) ;  that  is,  as  ABODE  X 
AF,  to  abcde  X  af.  But  since  the  prisms  are  similar,  the  bases 
are  similar  figures,  and  ire  to  each  other  as  the  squares  of 


BOOK   VIII. 


139 


B      C 

(he  i  homologous  sides ;  that  is,  as  AB2  to  ab*.     Therefore, 
we  have 

Solid  FD  :  solid  fd  :  :  AB'xAF  :  atfxaf. 
But  since  BF  and  bf  are  similar  figures,  their  homologous 
sides  are  proportional ;  that  is, 

AB   :  ab  :  :  AF  :  of, 
whence  (Prop.  X.,  B.  IL), 

AB2  :  ab*  :  :  AF3 :  of*. 
Also  AF  :  af  :  :  AF  :  of. 

Therefore  (Prop.  XL,  B.  IL), 

AB2x AF  :  atfxaf:  :  AF3  :  af3  :  :  AB3  :  ab9. 
Hence  (Prop.  IV.,  B.  IL),  we  have 

Solid  FD  :  solid  fd :  :  AB3  :  ab3  :  :  AF3  :  af. 
Therefore,  similar  prisms,  &c. 


PROPOSITION   XIII.       THEOREM. 


If  a  pyramid  be  cut  by  a  plane  parallel  to  its  base, 

1st.  The  edges  and  the  altitude  will  be  divided  proportionally. 

2d.   The  section  will  be  a  polygon  similar  to  the  base. 


Let  A-BCDEF  be  a  pyramid  cut  by  a 
plane  bcdef  parallel  to  its  base,  and  let 
AH  be  its  altitude ;  then  will  the  edges 
AB,  AC,  AD,  &c.,  with  the  altitude  AH, 
be  divided  proportionally  in  b,  c,  d,  e,  f, 
h ;  and  the  section  bcdef  will  be  similar  to 
BCDEF. 

First.  Since  the  planes   FBC,  fbc  are 
parallel,  their  sections  FB,  fb  with  a  third 
plane   AFB   are  parallel   (Prop.  XIL,  B. 
VII.) ;  therefore  the  triangles  AFB,  Afb 
are  similar,  and  we  have  the  proportion 
AF:  A/::  AB  :  A&. 
For  the  same  reason, 

AB  :  Ab  :  :  AC  :  Ac. 


d, 


140  GEOMETRY. 

and  so  for  the  other  edges.  Thei'efo:e,  tiie  edges  AB,  AC, 
&c.,  are  cut  proportionally  in  b,  c,  &c.  Also,  since  BH  and 
bh  are  parallel,  we  have 

AH  :  Ah  :  :  AB  :  Ab. 

Secondly    Because  fb  is  parallel  to  FB,  be  to  BC,  cd  x>  CD 
&c.,  the  angle  fbc  is  equal  to  FBC  (Prop.  XV.,  B.  VII.),  tht 
angle  bed  is  equal  to  BCD,  and  so  on.     Moreover,  since  the 
triangles  AFB,  Afb  are  similar,  we  have 
FB  :fb  :  :  AB  •  Ab. 
And  because  the  triangles  ABC,  Abe  are  similar,  we  have 

AB  :  Ab  :  :  BC  :  be. 
Therefore,  by  equality  of  ratios  (Prop.  IV.,  B.  II.), 

FB  :fb  :  :  BC  :  be. 
For  the  same  reason, 

BC  :  be  :  :  CD  :  cd,  and  so  on. 

Therefore  the  polygons  BCDEF,  bcdef  have  their  angles 
equal,  each  to  each,  and  their  homologous  sides  proportional ; 
hence  they  are  similar.  Therefore,  if  a  pyramid,  &c. 

Cor.  1.  If  two  pyramids,  having  the  same  altitude,  and  their 
bases  situated  in  the  same  plane,  are  cut  by  a  plane  parallel  to 
their  bases,  the  sections  will  be  to  each  other  as  the  bases. 

Let  A-BCDEF,  A-MNO 
be  two  pyramids  having 
the  same  altitude,  and  their 
^ases  situated  in  the  same 
plane  ;  if  these  pyramids 
are  cut  by  a  plane  parallel 
to  the  bases,  the  sections 
bcdef,  mno  will  be  to  each 
other  as  the  bases  BCDEF, 
MNO. 

For,  since   the   polygons 
BCDEF,  bcdef  are  similar, 
their  surfaces  are  as  the  squares  of  the  homologous  sides  BC 
be  (Prop.  XXVI.,  B.  IV.).     But,  by  the  preceding  Proposition 

BC  :  be  :  :  AB  :  Ab. 

Therefore,         BCDEF  :  bcdef:  :  AB2  :  Ab\ 
For  the  same  reason, 

MNO  :  mno  :  :  AM2  :  Am1. 
But  since  bcdef  and  mno  are  in  the  same  plane,  we  have 

AB  :  Ab  :  :  AM  :  Am  (Prop.  XVI.,  B.  VII.) ; 
consequently,     BCDEF  :  bcdef :  :  MNO  :  mno. 

Cor.  2.  If  the  bases  BCDEF,  MNO  are  equivalent,  th« 
sections  bcdef,  mno  will  also  be  equivalent. 


BOOK    VIII. 


PROPOSITION    XIV.       THEOREM. 


141 


The  convex  surface  of  a  regular  pyramid,  is  equal  to  the 
verimeter  of  its  base,  multiplied  by  half  the  slant  height 

Let  A-BDE  be  a  regular  pyramid,  whose 
base  is  the  polygon  BCDEF,  and  its  slant 
height  AH ;  then  will  its  convex  surface  be 
equal  to  the  perimeter  BC+CD-j-DE,  &c., 
multiplied  b>  half  of  AH. 

The  triangles  AFB,  ABC,  ACD,  &o.,  are 
all  equal  for  the  sides  FB,  BC,  CD,  &c.,  are 
all  equal,  (Def.  13) ;  and  since  the  oblique 
lines  AF,  AB,  AC,  &c.,  are  all  at  equal  dis- 
tances from  the  perpendicular,  they  are 
equal  to  each  other  (Prop.  V.,  B.  VII.). 
Hence  the  altitudes  of  these  several  triangles 
are  equal.  But  the  area  of  the  triangle  AFB  is  equal  to  FB, 
multiplied  by  half  of  AH  ;  and  the  same  is  true  of  the  other 
triangles  ABC,  ACD,  &c.  Hence  the  sum  of  the  triangles  is 
equal  to  the  sum  of  the  bases  FB,  BC,  CD,  DE,  EF,  multiplied 
by  half  the  common  altitude  AH ;  that  is,  the  convax  surface 
of  the  pyramid  is  equal  to  the  perimeter  of  its  base,  multiplied 
by  half  the  slant  height. 

Cor.  1.  The  convex  surface  of  a  frustum  of  a  regular 
pyramid  is  equal  to  the  sum  of  the  perimeters  of  its  two  bases, 
multiplied  by  half  its  slant  height. 

Each  side  of  a  frustum  of  a  regular  pyramid,  as  FB£/,  is  a 
trapezoid  (Prop.  XIII.)..  Now  the  area  of  this  trapezoid  is 
equal  to  the  sum  of  its  parallel  sides  FB,  fb,  multiplied  by 
half  its  altitude  HA  (Prop.  VII.,  B.  IV.).  But  the  altitude 
of  each  of  these  trapezoids  is  the  same  ;  therefore  the  area  of 
all  the  trapezoids,  or  the  convex  surface  of  the  frustum,  is 
equal  to  the  sum  of  the  perimeters  of  the  two  bases,  multiplied 
by  half  the  slant  height. 

Cor.  2.  If  the  frustum  is  cut  by  a  plane,  parallel  to  the  bases, 
and  at  equal  distances  from  them,  this  plane  must  bisect  the 
edges  B&,  Cc,  &c.  (Prop.  XVI.,  B.  IV.) ;  and  the  area  of  each 
trapezoid  is  equal  to  its  altitude,  multiplied  by  the  line  which 
joins  the  middle  points  of  its  two  inclined  sides  (Prop.  VII. , 
Cor.,  B.  IV.).  Hence  the  convex  surface  of  a  frustum  of  a 
pyramid  is  equal  to  its  slant  height,  multiplied  by  the 
perimeter  of  a  section  at  equal  distances  between  the  two 
bases. 


142 


GEOMETRY, 


PROPOSITION    XV.       THEOREM. 


Triangular  pyramids,  having  equivalent  bases  and  equal 
Etudes, are  equivalent. 


Let  A-BCD,  a-bcd  be  two  triangular  pyramids  having 
equivalent  bases  BCD,  bed,  supposed  to  be  situated  in  the 
same  plane,  and  having  the  common  altitude  TB ;  then  will 
the  pyramid  A-BCD  be  equivalent  to  the  pyramid  a-bcd. 

For,  if  they  are  not  equivalent,  let  the  pyramid  A-BCD 
exceed  the  pyramid  a-bcd  by  a  prism  whose  base  is  BCD 
and  altitude  BX. 

Divide  the  altitude  BT  into  equal  parts,  each  less  than 
BX  ;  and  through  the  several  points  of  division, let  planes  be 
made  to  pass  parallel  to  the  base  BCD,  making  the  sections 
EFG,  efg  equivalent  to  each  other  (Prop.  XIIL,  Cor.  2)  - 
also,  HlK  equivalent  to  hik,  &c. 

From  the  point  C,  draw  the  straight  line  CR  parallel  to 
BE,  meeting  EF  produced  in  R ;  and  from  D  draw  DS  par- 
alle.  to  BE,  meeting  EG  in  S.  Join  RS,  and  it  is  plain  that 
the  solid  BCD-ERS  is  a  prism  lying  partly  without  the  pyr 
amid.  In  the  same  manner,  upon  the  triangles  EFG,  HIK, 
&c.,  taken  as  bases,  construct  exterior  prisms,  having  for 
edges  the  parts  EH,  HL,  &c.,  of  the  line  AB.  In  like  man 
ner,  on  the  bases  efg,  hik,  Imn,  &c.,  in  the  second  pyramid, 
construct  interior  prisms,  having  for  edges  the  corresponding 
oarts  of  ab.  It  is  plain  that  the  sum  of  all  the  exterior  prisms 


BOOK    VII  f.  143 

ut  tne  pyramid  A-BCD  is  greater  than  this  pyi&mid;  and, 
also,  that  the  sum  of  all  the  interior  ^risms  of  the  pyramid 
a— bed  is  smaller  than  this  pyramid.  Hence  the  difference 
between  the  sum  of  all  the  exterior  prisms,  and  the  sum  of 
all  the  interior  ones,  must  be  greater  than  the  difference  be 
tween  the  two  pyramids  themselves. 

Now,  beginning  with  the  bases  BCD,  bed,  the  second  ex 
terior  prism  EFG-H  is  equivalent  to  the  first  interior  prism 
efg—b,  because  their  bases  are  equivalent,  and  they  have  the 
same  altitude.  For  the  same  reason,  the  third  exterior  prism 
HIK-L  and  the  second  interior  prism  hik-e  are  equivalent ; 
the  fourth  exterior  and  the  third  interior ;  and  so  on,  to  the 
last  in  each  series.  Hence  all  the  exterior  prisms  of  the  pyr- 
amid A— BCD,  excepting  the  first  prism  BCD— E,  have  equiv- 
.ent  corresponding  ones  in  the  interior  prisms  of  the  pyramid 
a-bcd.  Therefore  the  prism  BCD-E  is  the  difference  be- 
tween the  sum  of  all  the  exterior  prisms  of  the  pyramid 
A-BCD,  and  the  sum  of  all  the  interior  prisms  of  the  pyr- 
amid a-bcd.  But  the  difference  between  these  two  sets  of 
prisms  has  been  proved  to  be  greater  than  that  of  the  two 
pyramids  ;  hence  the  prism  BCD-E  is  greater  than  the  prism 
BCD— X  ;  which  is  impossible,  for  they  have  the  ^same  base 
BCD,  and  the  altitude  of  the  first,  is  less  than  BX,  the  altitude 
of  the  second.  Hence  the  pyramids  A-BCD,  a-bcd  are  not 
unequal ;  that  is,  they  are  equivalent  to  each  other.  There- 
fore, triangular  pyramids,  &c. 


PROPOSITION   XVI.       THEOREM. 

Every  triangular  pyramid  is  the  third  part  of  a  tnangulai 
prism  having  the  same  base  and  the  same  altitude. 

Let  E-ABC  be  a  triangular  pyramid, 
and  ABC-DEF  a  triangular  prism  hav- 
ing the  same  base  and  the  same  altitude  ; 
then  will  the  pyramid  be  one  third  of  the 
prism. 

Crrf  off  from  the  prism  the  pyramid 
E-AiiC  by  the  plane  EAC  ;  there  will  re- 
main the  solid  E-ACFD,  which  may  be 
considered  as  a  quadrangular  pyramid 
whose  vertex  is  E,  and  whose  base  is  the 
paiaJelogram  ACFD.  Draw  the  diago- 
nal CD,  and  through  the  points  C,  D,  E  pass  a  plane,  dividing 
.he  quadrangular  pyramid  into  two  triangular  ones  E-ACD 
E-CFD.  Then,  because  ACFD  is  a  oarallelo^ram,  of  whin? 


144 


GEOMETRY. 


CD  is  the  diagonal,  the  triangle  ACD  is 
equal  to  the  triangle  CDF.  Therefore 
the  pyramid,  whose  base  is  the  triangle 
ACD,  and  vertex  the  point  E,  is  equiva- 
lent to  the  pyramid  whose  base  is  the  tri- 
angle CDF,  and  vertex  the  point  E.  But 
the  latter  pyramid  is  equivalent  to  the 
pyramid  E-ABC  for  they  have  equal 
bases,  viz.,  the  triangles  ABC,  DEF,  and 
the  same  altitude,  viz.,  the  altitude  of  the 
prism  ABC-DEF.  Therefore  the  three 
pyramids  E-ABC,  E-ACD,  E-CDF,  are  equivalent  to  each 
other,  and  they  compose  the  whole  prism  ABC-DEF  ;  hence 
the  pyramid  E-ABC  is  the  third  part  of  the  prism  which 
has  the  same  base  and  the  same  altitude. 

Cor.  The  solidity  of  a  triangular  pyramid  is  measured  hv 
the  product  of  its  base  by  one  third  of  its  altitude. 


PROPOSITION   XVII.       THEOREM. 

The  solidity  of  every  pyramid  is  measured  by  the  product  oj 
tts  base  by  one  third  of  its  altitude. 

Let  A-BCDEF  be  any  pyramid,  whose 
base  is  the  polygon  BCDEF,  and  altitude 
AH ;  then  will  the  solidity  of  the  pyramid 
be  measured  by  BCDEF  x^AE. 

Divide  the  polygon  BCDEF  into  triangles 
by  the  diagonals  CF,  DF ;  and  let  planes 
pass  through  these  lines  and  the  vertex  A ; 
they  will  divide  the  polygonal  pyramid 
A-BCDEF  into  triangular  pyramids,  all 
having  the  same  altitude  AH.  But  each  of 
these  pyramids  is  measured  by  the  product 
of  its  base  by  one  third  of  its  altitude  (Prop. 
XVI.,  Cor.) ;  hence  the  sum  of  the  triangular  pyramids,  or 
the  polygonal  pyramid  A-BCDEF,  will  be  measured  by  ihe 
gum  of  ..the  triangles  BCF,  CDF,  DEF,  or  the  polygon 
BCDEF,  multiplied  by  one  third  of  AH.  Therefore  every 
pyramid  is  measured  by  the  product  of  its  base  by  one  third 
of  its  altitude. 

Cor.  1.  Every  pyramid  is  one  third  of  a  prism  having  the 
§ame  base  and  altitude. 

Cor.  2.  Pyramids  of  the  same  altitude  are  to  each  other 
as  their  bases ;  pyramids  of  the  same  base  are  to  each  other 


BOOK    VII!. 


145 


as  their  altitudes  ;  and  pyramids  generally  are  to  each  other 
as  the  products  of  their  bases  by  their  altitudes. 

Co/  3.  Similar  pyramids  are  to  each  other  as  the  cubes 
of  their  homologous  edges. 

Scholium.  The  solidity  of  any  polyedron  may  be  found 
by  dividing  it  into  pyramids,  by  planes  passing  through  its 
vertices. 


PROPOSITION    XVIII.       THEOREM. 

A  frustum  of  a  pyramid  is  equivalent  to  the  sum  of  tnree 
pyramids,  having  the  same  altitude  as  the  frustum,  and  whose 
bases  are  the  lower  base  of  the  frustum,  its  upper  base,  and  a 
mean  proportional  between  them. 

Case  first.  When  the  base  of  the  frustum  is  a  triangle. 

Let  ABC-DEF  be  a  frustum  of  a  tri- 
angular pyramid.  If  a  plane  be  made  to 
pass  through  the  points  A,  C,  E,  it  will 
cut  off  the  pyramid  E-ABC,  whose  alti- 
tude is  the  altitude  of  the  frustum,  and 
its  base  is  ABC,  the  lower  base  of  the 
frustum. 

Pass  another  plane  through  the  points 
C,  D,  E ;  it  will  cut  off  the  pyramid 
C-DEF,  whose  altitude  is  that  of  the 
frustum,  and  its  base  is  DEF,  the  upper 
base  of  the  frustum. 

To  find  the  magnitude  of  the  remaining  pyramid  E-ACD, 
draw  EG  parallel  to  AD  ;  join  CG,  DG.     Then,  because  the 
two  triangles  AGC,  DEF  have  the  angles  at  A  and  D  equal 
to  each  other,  we  have  (Prop.  XXIIL,  B.  IV.) 
AGC  :  DEF  :  :  AG X AC  :  DE xDF, 

:  :  AC  :  DF,  because  AG  is  equal  to  DE. 
Also  (Prop.  VI.,  Cor.  1,  B.  IV.), 

ACB  :  ACG  :  :  AB  :  AG  or  DE. 

But,  because  the  triangles  ABC,  DEF  are  similar  (Prop. 
XIII.),  we  have 

AB  :  DE  :  :  AC  :  DF. 
Therefore  (Prop.  IV.,  B.  II.), 

ACB  :  ACG  :  :  ACG  :  DEF ; 

that  is,  the  triangle  ACG  is  a  mean  proportional  between 
ACB  and  DEF,  the  two  bases  of  the  frustum. 

Now  the  pyramid  E-ACD  is  equivalent  to  the  pyramid 
G-ACD,  because  it  hat  the  same  base  and  the  same  altitude  • 
for  EG  is  parallel  to  AD,  and,  consequently,  parallel  to  the 

G 


146 


GEOMETRY 


plane  ACD.  But  the  pyramid  G-ACD  has  the  same  altitude 
as  the  frustum,  and  its  base  ACG  is  a  mean  proportional  be 
tween  the  two  bases  of  the  frustum. 

Case  second.  When  the  base  of  the  frustum  is  any  polys  oa 
Let  ECVEF-bcdef  be   a  A  G  " 

frustum  of  any  pyramid.   v°* 

Let  G-HIK  be  a  trian- 
gular pyramid  having  the 
same  altitude  and  an  equiv- 
alent base  with  the  pyramid 
A-BCDEF,  and  from  it  let 
a  frustum  HIK— /M&  be  cut 
off,  having  the  same  altitude 
with  the  frustum  BCDEF-  C 

bcdef.  The  entire  pyramids  are  equivalent  (Prop.  XVII.; 
and  the  small  pyramids  A-bcdef,  G-hik  are  also  equivalent, 
for  their  altitudes  are  -equal,  and  their  bases  are  equivalent 
(Prop.  XIII.,  Cor.  2).  Hence  the  two  frustums  are  equiva- 
lent, and  they  have  the  same  altitude,  with  equivalent  bases. 
But  the  frustum  HIK.-hik  has  been  proved  to  be  equivalent  to 
the  sum  of  three  pyramids,  each  having  the  same  altitude  as 
the  frustum,  and  whose  bases  are  the  lower  base  of  the  frus- 
tum, its  upper  base,  and  a  mean  proportional  between  them 
Hence  the  same  must  be  true  of  the  frustum  of  any  pyramid 
Therefore,  a  frustum  of  a  pyramid,  &c. 


PROPOSITION    XIX.       THEOREM. 

There  can  be  but  Jive  regular  polyedrons. 

Since  the  faces  of  a  regular  polyedron  are  regular  poly 
gons,  they  must  consist  of  equilateral  triangles,  of  squares,  of 
regular  pentagons,  or  polygons  of  a  greater  number  of  sides. 

First.  If  the  faces  are  equilateral  triangles,  each  solid  an- 
le  of  the  polyedron  may  be  contained  by  three  of  these  trr 


angles,  forming  the  tetraedron ;  or  by  four,  forming  the  0c- 
taedron ;  or  by  five,  forming  the  icosaedron. 

No  other  regular  polyedron  can  be  formed  with  equilat 
eral  triangles ;  for  six  angles  of  these  triangles  amount  t« 


BOOK  VIII 


141 


lour  right  angles,  and  can  not  form  a  solid  angle 
(Prop.  XVIIL,  B.  VII.). 

Secondly.  If  the  faces  are  squares,  their  an- 
gles may  be  united  three  and  three,  forming 
the  hexaedron,  or  cube. 

Four  angles  of  squares  amount  to  four  right 
angles,  and  can  not  form  a  solid  angle. 

Thirdly.  If  the  faces  are  regular  penta- 
gons, their  angles  may  be  united  three  and 
three,  forming  the  regular  dodedhedron.  Four 
angles  of  a  regular  pentagon,  are  greater 
than  four  right  angles,  and  can  not  form  a 
solid  angle. 

Fourthly.  A  regular  polyedron  can  not  be 
formed  with  regular  hexagons,  for  three  angles  of  a  regular 
hexagon  amount  to  four  right  angles.  Three  angles  of  a 
regular  heptagon  amount  to  more  than  four  right  angles ; 
and  the  same  is  true  of  any  polygon  having  a  greater  number 
of  sides. 

Hence  there  can  be  but  five  regular  jolyedrons;  three 
formed  with  equilateral  triangles,  one  with  squares,  and  one 
with  pentagons 


148  GEOMETRY. 


BOOK  IX. 

SPHERICAL  GEOMETRY 
Definitions. 

1.  A  sphere  is  a  solid  bounded  by  a  curved  surface,  all  the 

of  which  are  equally  distant  from  a  point  within,  call- 
ed the  center. 

The  sphere  may  be  conceived  to  be  de- 
scribed by  the  revolution  of  a  semicircle 
ADB,  about  its  diameter  AB,  which  re- 
mains unmoved. 

2.  The  radius  of  a  sphere,  is  a  straight 
line  drawn  from  the  center  to  any  point  of 
the  surface.    The  diameter,  or  axis,  is  a  line 
passing  through  the  center,  and  terminated 
each  way  by  the  surface. 

All  the  radii  of  a  sphere  are  equal ;  all  the  diameters  are 
also  equal,  and  each  double  of  the  radius. 

3.  It  will  be  shown  (Prop.  I.),  that  every  section  of  a 
sphere  made  by  a  plane  is  a  circle.     A  great  circle  is  a  sec- 
tion made  by  a  plane  which  passes  through  the  center  of  the 
sphere.     Any  other  section  made  by  a  plane  is  called  a  small 
circle. 

4.  A  plane  touches  a  sphere, when  it  meets  the  sphere,  but 
being  produced,  does  not  cut  it. 

5.  The  pole  of  a  circle  of  a  sphere,  is  a  point  in  the  surface 
equally  distant  from  every  point  in  the  circumference  of 
this  circle.     It  will  be  shown  (Prop.  V.),  that  every  circle, 
whether  great  or  small,  has  two  poles. 

6.  A  spherical  triangle  is  a  part  of  the  sur- 
face of  a  sphere,  bounded  by  three  arcs  of 
great  circles,  each  of  which  is  less  than  a  semi- 
circumference.     These  arcs  are  called  the  sides 
of  the   triangle ;   and  the  angles  which  their 
planes  make  with  each  other,  are  the  angles 
of  the  triangle. 

"i    A  spherical  triangle  is  called  right-angled,  isosceles  or 
tquiiateral,  in  the  SJMne  cases  as  a  plane  triangle. 


BOOK    IX. 


:.  A  spherical  polygon  is  a  part  of  Ine  sur- 
face of  a  sphere  bounded  by  several  arcs  of 
great  circles. 


9.  A  lune  is  a  part  of  the  surface  of  a  sphere  in- 
cluded between  the  halves  of  two  great  circles. 

10.  A  spherical  wedge,  or  ungula,  is  that  portion 
of  the  sphere  included  between  the  same  semicir- 
cles, and  has  the  lune  for  its  base. 


11.  A  spherical  pyramid  is  a  portion  of  the 
sphere  included  between  the  planes  of  a  solid 
angle,  whose  vertex  is  at  the  center.  The  base 
of  the  pyramid  is  the  spherical  polygon  inter- 
cepted by  those  planes. 


12.  A  zone  is  a  part  of  the  surface  of  a 
sphere  included  between  two  parallel  planes. 

13.  A  spherical  segment  is  a  portion  of  the 
sphere  included  between  two  parallel  planes. 

14.  The  bases  of  the  segment  are  the  sec- 
tions of  the  sphere ;  the  altitude  of  the  seg- 
ment, or  zone,  is  the  distance  between  the 

sections.     One  of  the  two  planes  may  touch  the  sphere,  ir». 
which  case  the  segment  has  but  one  base. 

15.  A  spherical  sector  is  a  solid  de- 
scribed by  the  revolution  of  a  circular 
sector,   in   the    same    manner    as    the 
sphere  is  described  by  the  revolution 
of  a  semicircle. 

While  the  semicircle  ADB,  revolving 
round  its  diameter  AB,  describes  a 
sphere,  every  circular  sector,  as  ACE 
or  ECD,  describes  a  spherical  sector. 

16.  Two  angles  which  are  together 
equal  to  two  right  angles ;  or  two  arcs 

which  are  together  equal  to  a  semicircum'erer.ce,  are  called 
the  supplements  of  each  other. 


E 


HO  GEOMETRY. 

• 
PROPOSITION  I.       THEOREM. 

'Ewy  section  of  a  sphere,  made  by  i  plane,  is  a  circle 

Let  ABD  be  a  section,  made  by  a 
plane,  in  a  sphere  whose  center  is  C. 
From  the  point  C  draw  CE  perpendicu- 
lar  to  the  plane  ABD  ;  and  draw  lines 
CA,  CB,  CD,  &c.,  to  different  points  of 
the  curve  ABD  which  bounds  the  sec- 
tion. 

The  oblique  lines  CA,  CB,  CD  are 
equal,  because  they  are  radii  of  the 
sphere;  therefore  they  are  equally  distant  from  the  perpen- 
dicular CE  (Prop.  V.,  Cor.,  B.  VII.).  Hence  all  the  lines 
EA,  EB,  ED  are  equal ;  and,  consequently,  the  section  ABD 
is  a  circle,  of  which  E  is  the  center.  Therefore,  every  sec- 
tion, &c. 

Cor.  1.  If  the  section  passes  through  the  center  of  the 
sphere,  its  radius  will  be  the  radius  of  the  sphere ;  hence  al1 
great  circles  of  a  sphere  are  equal  to  each  other. 

Cor.  2.  Two  great  circles  always  bisect  each  other ;  for, 
since  they  have  the  same  center,  their  common  section  is  a 
diameter  of  both,  and  therefore  bisects  both. 

Cor.  3.  Every  great  circle  divides  the  sphere  and  its  sur- 
face into  two  equal  parts.  For  if  the  two  parts  are  separated 
and  applied  to  each  other,  base  to  base,  with  their  convexities 
turned  the  same  way,  the  two  surfaces  must  coincide  ;  oth- 
erwise there  would  be  points  in  these  surfaces  unequally  dis- 
tant from  the  center. 

Cor.  4.  The  center  of  a  small  circle,  and  that  of  the  sphere, 
are  in  a  straight  line  perpendicular  to  the  -plane  of  the  small 
circle. 

Cor.  5.  The  circle  which  is  furthest  from  the  center  is  the 
least ;  for  the  greater  the  distance  CE,  the  less  is  the  chord 
AB,  which  is  the  diameter  of  the  small  circle  ABD. 

Cor.  6.  An  arc  of  a  great  circle  may  be  made  to  pass 
through  any  two  points  on  the  surface  of  a  sphere  ;  for  the 
two  given  points,  together  with  the  center  of  the  sphere, 
make  three  points  which  are  necessary  to  determine  the  posi- 
tion of  a  plane.  If,  however,  the  two  given  points  were  sit- 
uated at  the  extremities  of  a  diameter,  these  two  points  and 
the  center  would  then  be  in  one  straight  line,  and  any  mini 
ber  of  great  oi'des  might  be  made  to  pass  through  them. 


BOOK    IX.  151 


PROPOSITION   II.       THEOREM. 

Any  tw<)  sides  of  a  spherical  triangle  are  together  great- 
er than  the  third. 

Let  ABC  be  a  spherical  triangle ;  any 
two  sides  as,  AB,  BC,  are  together  greater 
than  the  third  side  AC. 

Let  D  be  the  center  of  the  sphere  ;  and 
^oin  AD,  BD,  CD.     Conceive  the  planes 
ADB,  BDC,  CDA  to  be  drawn,  forming  a 
solid  angle  at  D.     The  angles  ADB,  BDC, 
CDA  will  be  measured  by  AB,  BC,  CA, 
he  sides  of  the  spherical   triangle.     But 
when  a  solid  angle  is  formed  by  three  plane  angles,  the  sum 
of  any  two  of  them  is  greater  than  the  third  (Prop.  XVII.,  B. 
VII.) ;  hence  any  two  of  the  arcs  AB,  BC,  CA  must  b 
greater  than  the  third.     Therefore,  any  two  sides,  &c. 


PROPOSITION  III.       THEOREM. 

The  shortest  path  from  one  point  to  another  on  the  surface, 
:-f  a  sphere,  is  the  arc  of  a  great  circle  joining  the  two  given 
points. 

Let  A  and  B  be  any  two  points  on  the  surface  of 
a  sphere,  and  let  ADB  be  the  arc  of  a  great  circle 
which  joins  them ;  then  will  the  line  ADB  be  the 
shortest  path  from  A  to  B  on  the  surface  of  the 
sphere. 

For,  if  possible,  let  the  shortest  path  from  A  to  B 
pass  through  C,  a  point  situated  out  of  the  arc  of  a. 
great  circle  ADB.  Draw  AC,  CB,  arcs  of  great 
circles,  and  take  BD  equal  to  BC. 

By  the  preceding  theorem,  the  arc  ADB  is  less  than  AC-f- 
CB.  Subtracting  the  equal  arcs  BD  and  BC,  there  will  re- 
main AD  less  than  AC.  Now  the  shortest  path  from  B  to  C, 
whether  it  be  an  arc  of  a  great  circle,  or  some  other  line,  is 
equal  to  the  shortest  path  from  B  to  D ;  for,  by  revolving 
BC  around  B,  the  point  C  may  be  made  to  coincide  with  D, 
and  thus  the  shortest  path  from  B  to  C  must  coincide  with 
the  shortest  path  from  B  to  D.  But  the  shortest  path  from 
A  to  B  was  supposed  to  pass  through  C  ;  hence  the  shortest 
path  from  A  to  C,  can  not  be  greater  than  the  shortest  path 
A  to  D. 


152 


GEOMETRY 


Now  the  arc  AD  has  been  proved  to  be  less  than  AC ;  and 
therefore  if  AC  be  revolved  about  A  until  the  point  C  falls 
on  the  arc  ADB,  the  point  C  will  fall  between  D  and  B. 
Hence  the  shortest  path  from  C  to  A  must  be  greater  than 
the  shortest  path  from  D  to  A ;  but  it  has  just  been  proved 
not  to  be  greater,  which  is  absurd.  Consequently,  no  poin, 
of  the  shortest  path  from  A  to  B,  can  be  out  of  the  arc  of  9 
great  circle  ADB.  Therefore,  the  shortest  path,  &c. 


PROPOSITION  IV.       THEOREM. 

The  sum  of  the  sides  of  a  spherical  polygon,  is  less  than  the 
4  circumference  of  a  great  circle. 

Let  ABCD  be  any  spherical  polygon ; 
then  will  the  sum  of  the  sides  AB,  BC,  CD, 
DA  be  less  than  the  circumfeience  of  a 
great  circle. 

Let  E  be  the  center  of  the  sphere,  and 
join  AE,  BE,  CE,  DE.  The  solid  angle 
at  E  is  contained  by  the  plane  angles  AEB, 
BEC,  CED,  DEA,  which  together  are  less 
than  four  right  angles  (Prop.  XVIIL,  B. 
VII.) .  Hence  the  sides  AB,  BC,  CD,  DA, 
which  are  the  measures  of  these  angles,  are 
together  less  than  four  quadrants  described  with  the  radius 
AE ;  that  is,  than  the  circumfeience- of  a  great  circle 
Therefore,  the  sum  of  the  sides,  &c. 


PROPOSITION    V.       THEOREM. 

The  extremities  of  a  diameter  of  a  sphere?  are  the  poles  of  aH 
circles  perpendicular  to  that  diameter. 

Let  AB  be  a  diameter  perpendicu- 
ar  to  CDE,  a  great  circle  of  a  sphere, 
and  also  to  the  small  circle  FGH ; 
then  will  A  and  B,  the  extremities  of 
the  diameter,  be  the  poles  of  both 
these  circles. 

For,  because  AB  is  perpendicular 
to  the  plane  CDE,  it  is  perpendicular 
to  every  straight  line  CI,  DI,  El,  &c., 
drawn  through  its  foot  in  the  plane  ; 
nence  all  the  arcs  AC,  AD,  AE,  &c.,  are  quarters  of  the 


BOOK    IX.  153 

cumference.  So,  also,  the  arcs  BC,  BD,  BE,  &c.,  are  quar- 
ters of  the  circumference ;  hence  the  points  A  and  B  are 
each  equally  distant  from  all  the  points  of  the  circumference 
CDE  ;  they  are,  therefore,  the  poles  of  that  circumference 
(Def.  5). 

Secondly.  Because  the  radius  AI  is  perpendicular  to  the 
plane  of  the  circle  FGH,  it  passes  through  K,  the  center  of 
that  circle  (Prop.  I.,  Cor.  4).  Hence,  if  we  draw  the  oblique 
lines  AF,  AG,  AH,  these  lines  will  be  equally  distant  from 
the  perpendicular  AK,  and  will  be  equal  to  each  other  (Prop. 
V.,  B.  VIL).  But  since  the  chords  AF,  AG,  AH  are  equal, 
the  arcs  are  equal ;  hence  the  point  A  is  a  pole  of  the  small 
circle  FGH ;  and  in  the  same  manner  it  may  be  proved  that 
B  is  the  other  pole. 

Cor.  1.  The  arc  of  a  great  circle  AD,  drawn  from  the  pole 
to  the  circumference  of  another  great  circle  CDE,  is  a  qua- 
drant ;  and  this  quadrant  is  perpendicular  to  the  arc  CD. 
For,  because  AI  is  perpendicular  to  the  plane  GDI,  every 
plane  ADB  which  passes  through  the  line  AI  is  perpendicu 
far  to  the  plane  GDI  (Prop.  VI.,  B.  VII.) ;  therefore  the  an 
gle  contained  by  these  planes,  or  the  angle  ADC  (Def.  6),  is 
a  right  angle. 

Cor.  2.  If  it  is  required  to  find  the  pole  of  the  arc  CD, 
draw  the  indefinite  arc  DA  perpendicular  to  CD,  and  take  DA 
equal  to  a  quadrant ;  the  point  A  will  be  one  of  the  poles  of 
the  arc  CD.  Or,  at  each  of  the  extremities  C  and  D,  draw 
the  arcs  C  A  and  DA  perpendicular  to  CD  ;  the  point  of  inter 
section  of  these  arcs  will  be  the  pole  required. 

Cor.  3.  Conversely,  if  the  distance  of  the  point  A  from 
each  of  the  points  C  and  D  is  equal  to  a  quadrant,  the  point 
A  will  be  the  pole  of  the  arc  CD ;  and  the  angles  ACD, 
ADC  will  be  right  angles. 

For,  let  I  be  the  center  of  the  sphere,  and  draw  the  radii 
AI,  CI,  DI.  Because  the  angles  AIC,  AID  are  right  angles, 
the  line  AI  is  perpendicular  to  the  two  lines  CI,  DI ;  it  is, 
therefore,  perpendicular  to  their  plane  (Prop.  IV.,  B.  VIL). 
Hence  the  point  A  is  the  pole  of  the  arc  CD  (Prop.  V.)  ;  and 
therefore  the  angles  ACD,  ADC  are  right  angles  (Cor.  1). 

Scholium.  Circles  may  be  drawn  upon  the  surface  of  a 
sphere,  with  the  same  ease  as  upon  a  plane  surface.  Thus, 
by  revolving  the  arc  AF  around  the  point  A,  the  point  F  will 
describe  the  smail  circle  FGH ;  and  if  we  revolve  the  qua- 
drant AC  around  the  point  A,  the  extremity  C  will  describe 
the  great  circle  CDE. 

If  it  is  required  to  produce  the  arc  CD,  or  if  it  is  required 
to  draw  an  arc  of  a  great  circle  through  the  two  points  C 
and  D,  then  from  the  points  C  a^d  D  Hi  renters,  with  a  radius 


154  GEOMETRY 

equal  to  a  quadrant,  describe  two  arcs  intersecting  each 
other  .11  A.  The  point  A  will  be  the  pole  of  the  arc  CD ; 
and,  therefore,  if,  from  A  as  a  center,  with  a  radius  equal  to  a 
quadrant,  we  describe  a  circle  CDE,  it  will  be  a  great  circle 
passing  through  C  and  D. 

If  it  is  required  to  let  fall  a  perpendicular  from  any  point 
G  upon  the  arc  CD ;  produce  CD  to  L,  making  GL  equal  to 
a  quadrant ;  then  from  the  pole  L,  with  the  radius  GL,  de- 
scribe the  arc  GD ;  it  will  be  perpendicular  to  CD. 


PROPOSITION    VI.       THEOREM. 

A  plane,perpendicular  to  a  diameter  at  its  extremity,  touches 
the  sphere. 

Let  ADB  be  a  plane  perpendicular  A_ 
to  the  diameter  DC  at  its  extremity ; 
then    the    plane    ADB    touches    the 
sphere: 

Let  E  be  any  point  in  the  plane 
ADB,  and  join  DE,  CE.  Because  CD 
is  perpendicular  to  the  plane  ADB,  it 
is  perpendicular  to  the  line  AB  (Def. 
1,  B.  VII.) ;  hence  the  angle  CDE  is  a  right  angle,  and  the 
line  CE  is  greater  than  CD.  Consequently,  the  point  E  lies 
without  the  sphere.  Hence  the  plane  ADB  has  only  the  point 
D  in  common  with  the  sphere  ;  it  therefore  touches  the  sphere 
(Def.  4).  Therefore,  a  plane,  &c. 

Cor.  In  the  same  manner,  it  may  be  \  roved  that  two 
spheres  touch  each  other,  when  the  distance  between  their 
centers  is  equal  to  the  sum  or  difference  of  their  radii ;  in 
which  case,  the  centers  and  the  point  of  contact  lie  in  one 
straight  line. 


PROPOSITION    VII.       THEOREM. 

The  angle  formed  by  two  arcs  of  great  circles,  is  equal  to 
the  angle  formed  by  the  tangents  of  those  arcs  at  the  point  of 
their  intersection ;  and  is  measured  by  the  arc  of  a  great  cir- 
cle described  from  its  vertex  as  a  pole,  and  included  between  its 
sides. 

Let  BAD  be  an  angle  formed  by  two  arcs  of  great  circles; 
then  will  it  be  equal  to  the  an^le  EAF  formed  by  the  tan- 


OOOK    IX. 


155 


gents  of  these  arcs  at  the  point  A , 
and  it  is  measured  by  the  arc  DB  de- 
scribe 1  from  the  vertex  A  as  a  pole. 

For  the  tangent  AE,  drawn  in  the 
plane  of  the  arc  AB,  is  perpendicular 
to  the  radius  AC  (Prop.  IX.,  B.  III.) ; 
also,  the  tangent  AF,  drawn  in  the 
plane  of  the  arc  AD,  is  perpendicular 
to  the  same  radius  AC.  Hence  the 
angle  E  AF  is  equal  to  the  angle  of  the 
planes  ACB,  ACD  (Def.  4,  B.  VII.),  which  is  the  same  a 
that  of  the  arcs  AB,  AD. 

Also,  if  the  arcs  AB,  AD  are  each  equal  to  a  quadrant,  the 
lines  CB,  CD  will  be  perpendicular  to  AC,  and  the  angle 
BCD  .will  be  equal  to  the  angle  of  the  planes  ACB,  ACD ; 
hence  the  arc  BD  measures  the  angle  of  the  planes,  or  the 
angle  BAD. 

Cor.  1.  Angles  of  spherical  triangles  may  be  compared 
with  each  other  by  means  of  arcs  of  great  circles  described 
from  their  vertices  as  poles,  and  included  between  their, 
sides  ;  and  thus  an  angle  can  easily  be  made,  equal  to  a  given 
angle. 

Cor.  2.  If  two  arcs  of  great  circles  AC, 
DE  cut  each  other,  the  vertical  angles  ABE, 
DBC  are  equal ;  for  each  is  equal  to  the  an- 
gle formed  by  the  two  planes  ABC,  DBE. 
Also,  the  two  adjacent  angles  ABD,  DBC 
are  together  equal  to  two  right  angles. 


PROPOSITION   VIII.       THEOREM. 

If  from  the  vertices  of  a  given  spherical  triangle,  as  poles, 
arcs  of  great  circles  are  described,  a  second  triangle  is  formed, 
whose  vertices  are  poles  of  the  sides  of  the  given  triangle. 

Let  ABC  be  a  spherical  triangle ; 
and  from  the  points  A,  B,  C,  as  poles, 
let  great  circles  be  described  inter- 
secting each  other  in  D,  E,  and  F ; 
then  will  the  points  D,  E,  and  F  be 
the  poles  of  the  sides  of  the  triangle 
ABC. 

For,  because  the  point  A  is  the  pole 
of  the  arc  EF,  the  distance  from  A  to 
E  is  a  quadrant.  Also,  because  the 
point  C  is  the  pole  of  the  arc  DE,  the 


I 56  GEOMETRY 

distance  from  C  to  E  is  a  quadrant.  Hence  the  point  E  is  at 
a  quadrant's  distance  from  each  of  the  points  A  and  C ;  it  is, 
therefore,  the  pole  of  the  arc  AC  (Prop.  V.,  Cor.  3).  In  the 
same  manner,  it  may  be  proved  that  D  is  the  pole  of  the  arc 
BC,  and  F  the  pole  of  the  arc  AB. 

Scholium.  The  triangle  DEF  is  called  the  polar  triangle 
of  ABC  ;  and  so,  also,  ABC  is  the  polar  triangle  of  DEF. 

Several  different  triangles  might  be  formed  by  producing 
the  sides  DE,  EF,  DF ;  but  we  shall  confine  ourselves  to  the 
central  triangle,  of  which  the  vertex  D  is  on  the  same  side 
of  BC  with  the  vertex  A ;  E  is  on  the  same  side  of  AC 
with  the  vertex  B ;  and  F  is  on  the  same  side  of  AB  with 
the  vertex  C. 


PROPOSITION    IX.       THEOREM. 

The  sides  of  a  spherical  triangle,  are  the  supplements  of  the 
arcs  which  measure  the  angles  of  its  polai  triangle ;  and  con- 
versely. 

Let  DEF  be  a  spherical  triangle, 
ABC  its  polar  triangle  ;  then  will  the 
side  EF  be  the  supplement  of  the  arc 
which  measures  the  angle  A  ;  and 
the  side  BC  is  the  supplement  of  the 
arc  which  measures  the  angle  D. 

Produce  the  sides  AB,  AC,  if  ne- 
cessary, until  they  meet  EF  in  G  and 
H.  Then,  because  the  point  A  is  the 
pole  of  the  arc  GH,  the  angle  A  is 
measured  by  the  arc  GH  (Prop.  VII.). 
Also,  because  E  is  the  pole  of  the  arc  AH,  the  arc  EH  is  a 
quadrant ;  and,  because  F  is  the  pole  of  AG,  the  arc  FG  is  a 
quadrant.  Hence  EH  and  GF,  or  EF  and  GH,  are  together 
equal  to  a  semicircumference.  Therefore  EF  is  the  supple- 
ment of  GH,  which  measures  the  angle  A.  So,  also,  DF  is 
the  supplement  of  the  arc  which  measures  the  angle  B ;  and 
DE  is  the  supplement  of  the  arc  which  measures  the  angle  C, 

Conversely.  Because  the  point  D  is  the  pole  of  the  arc  BC, 
the  angle  D  is  measured  by  the  arc  IK.  Also,  because  C  is 
the  pole  of  the  arc  DE,  the  arc  1C  is  a  quadrant ;  and,  be- 
cause B  is  the  pole  of  the  arc  DF,  the  arc  BK  is  a  quadrant. 
Hence  1C  and  BK,  or  IK  and  BC,  are  together  equal  to  a 
semicircumference.  Therefore  BC  is  the  supplement  of  IK, 
which  measures  the  angle  D.  So,  also,  AC  is  the  supple- 
ment of  the  arc  which  measures  the  angle  E ;  and  AB  is  thf 
supplement  of  the  arc  which  measures  the  angle  F. 


BOOK    IX.  15" 


PROPOSITION    X.       THEOREM 

The  sum  of  the  angles  of  a  spherical  triangle , is  greater  than 
two,  and  less  than  six  right  angles. 

Let  A,  B,  and  C  be  the  angles  of  a  spherical  triangle. 
The  arcs  which  measure  the  angles  A,  B,  and  C,  together 
with  the  three  sides  of  the  po[ar  triangle,  are  equal  to  three 
semicircumferences  (Prop.  IX  ).  But  the  three  sides  of  the 
polar  triangle  are  less  than  two  semicircumferences  (Prop. 
IV.)  ;  hence  the  arcs  which  measure  the  angles  A,  B,  and  C 
are  greater  than  one  semicircumference ;  and,  therefore,  the 
angles  A,  B,  and  C  are  greater  than  two  right  angles. 

Also,  because  each  angle  of  a  spherical  triangle  is  less  than 
two  right  angles,  the  sum  of  the  three  angl-es  must  be  less 
than  six  right  angles. 

Cor.  A  spherical  triangle  may  have  two,  or  /\ 

even  three,  right  angles  ;    also  two,  or  even         /    \ 
three,  obtuse  angles.     If  a  triangle  have  three      /          \ 
right  angles,  each  of  its  sides  will  be  a  qua-    /  \ 

drant,  and  the  triangle  is  called  a  quadrantal  I  \ 

triangle.     The  quadrantal  triangle  is  contain-  ^^ ^ 

ed  eight  times  in  the  surface  of  the  sphere. 


PROPOSITION   XI.       THEOREM. 

If  two  triangles  on  equal  spheres  are  mutually  equilateral, 
they  are  mutually  equiangular. 

Let  ABC,  DEF  be  two  triangles  on  equal  spheres,  having 
the  sides  AB  equal  to  DE,  AC  to  DF,  and  BC  to  EF ;  then 
will  the  angles  also  be  equal,  each  to  each. 

D 


B  E 

Let  the  centers  of  the  spheres  be  G  and  H,  and  draw  the 
radii  GA,  GB,  GC,  HD,  HE,  HF.  A  solid  angle  may  be  con 
ceived  as  formed  at  G  bt  the  three  plane  angles  AGB,  AGU 


I  55  GEOMETRY. 

BGC;  and  another  solid  angle  at  H  by  the,  thiee  plane  an- 
gles DUE,  DHF,  EHF.  Then,  because  the  arcs  AB,  DE 
are  equal,  the  angles  AGB,  DHE,  which  are  measured  by 
these  arcs,  are  equal.  For  the  same  reason,  the  angles  AGO, 
DHF  are  equal^  to  each  other ;  and,  also,  BGC  equal  to  EHF 

D 


E 

Hence  G  and  H  are  two  solid  angles  contained  by  three  equal 
plane  angles  ;  therefore  the  planes  of  these  equal  angles  are 
equally  inclined  to  each  other  (Prop.  XIX.,  B.  VII.) .  That 
is,  the  angles  of  the  triangle  ABC  are  equal  to  those  of  the 
triangle  DEF,  viz.,  the  angle  ABC  to  the  angle  DEF,  BAG 
to  EDF,  and  ACB  to  DFE. 

Scholium.  It  should  be  observed  that  the  two  triangles 
ABC,  DEF  do  not  admit  of  superposition,  unless  the  three 
sides  are  similarly  situated  in  both  cases.  Triangles  which 
are  mutually  equilateral,  but  can  not  be  applied  to  each  othei 
so  as  to  coincide,  are  called  symmetrical  triangles. 

PROPOSITION    XII.       THEOREM. 

If  two  triangles  on  equal  spheres  are  mutually  equiangular 
Ihey  are  mutually  equilateral. 

Denote  by  A  and  B  two  spherical  triangles  which  are  mu- 
tually equiangular,  and  by  P  and  Q  their  polar  triangles. 

Since  the  sides  of  P  and  Q,  are  the  supplements  of  the  arcs 
which  measure  the  angles  of  A  and  B  (Prop.  IX.),  P  and 
Q,  must  be  mutually  equilateral.  Also,  because  P  and  Q  are 
mutually  equilateral,  they  must  be  mutually  equiangular 
(Prop.  XL).  But  the  sides  of  A  and  B  are  the  supplements 
of  the  arcs  which  measure  the  angles  of  P  and  Q ;  and; 
therefore,  A  and  B  are  mutually  equilateral. 


PROPOSITION   XIII.       THEOREM. 

If  two  triangles  on  equal  spheres  have  two  sides,  and  the  in 
eluded  angle  of  the  one,  equal  to  two  sides  and  the  included 
angle  of  the  other,  each  to  each,  their  third  sides  will  be  equal, 
and  their  other  angles  will  be  equal,  each  to  each. 


BOOK    IX.  159 

Let  ABC,  DEF  be  two  triangles,  having  the  side  AB  equal 
to  DE.  AC  equal  to  DF,  and  the  angle  BAG  equal  to  the  an 
gle  EDF ;  then  will  the  side  BC  be  equal  to  EF,  the  angle 
ABC  to  DEF,  and  ACB  to  DFE. 

If  the  equal  sides  in  the  two  triangles  are  similarly  sit- 
uated, the  triangle  ABC  may  be  applied  to  the  triangle  DEF 
in  the  same  manner  as  in  plane  tri-  . 
angles  (Prop.  VI.,  B.  I.) ;  and  the  two 
triangles  will  coincide  throughout. 
Therefore  all  the  parts  of  the  one  tri- 
angle, will  be  equal  to  the  correspond- 
ing parts  of  the  other  triangle. 

But  if  the  equal  sides  in  the  two  tri- 
angles are  not  similarly  situated,  then 
construct  the  triangle  DF'E  symmet- 
rical  with  DFE,  having  DF'  equal  to  " 
DF,  and  EF/  equal  to  EF.  The  two  triangles  DEF',  DEF, 
oeing  mutually  equilateral,  are  also  mutually  equiangular 
(Prop.  XL).  Now  the  triangle  ABC  may  be  applied  to  the 
triangle  DEF',  so  as  to  coincide  throughout ;  and  hence  all 
the  parts  of  the  one  triangle,  will  be  equal  to  the  correspond- 
ing parts  of  the  other  triangle.  Therefore  the  side  BC,  be- 
ing equal  to  EF',  is  also  equal  to  EF ;  the  ai^gle  ABC,  being 
equal  to  DEF',  is  also  equal  to  DEF ;  and  the  angle  ACB, 
being  equal  to  DF'E,  is  also  equal  to  DFE.  Therefore,  if 
two  triangles,  &c. 


PROPOSITION    XIV.       THEOREM. 

If  two  triangles  on  equal  spheres  have  two  angles,  and  the 
included  side  of  the  one,  equal  to  two  angles  and  the  included  side 
of  the  other,  each  to  each,  their  third  angles  will  be  equal  and 
their  other  sides  will  be  equal,  each  to  each. 

If  the  two  triangles  ABC,  DEF  A 
have  the  angle  BAG  equal  to  the  an- 
gle EDF,  the  angle  ABC  equal  to 
DEF,  and  the  included  side  AB  equal 
to  DE ;  the  triangle  ABC  can  be 
placed  upon  the  triangle  DEF,  or 
upon  its  symmetrical  triangle  DEF', 
so  as  to  coincide.  Hence  the  remain- 
ing parts  of  the  triangle  ABC,  will  be 
equal  to  the  remaining  parts  of  the  triangle  DEF;  that  is, 
the  side  A  1  will  be  equal  to  DF,  BC  to  EF,  and  the  angle 
ACB  to  the  angle  DFE  Therefore,  if  two  triangles,  &c 


160 


GEOMETRY. 


PROPOSITION   XV.       THEOREM. 

If  two  triangles  on  equal  spheres  are  mutually  equilateral, 
they  are  equivalent. 

Let  ABC,  DEF  be  two  triangles 
which  have  the  three  sides  of  the 
one,  equal  to  the  three  sides  of  the 
other,  each  to  each,  viz.,  AB  to 
DE,  AC  to  DF,  and  BC  to  EF ; 
then  will  the  triangle  ABC  be 
equivalent  to  the  triangle  DEF. 

Let  G  be  the  pole  of  the  small 
circle  passing  through  the  three 
points  A,  B,  C;  draw  the  arcs  GA,  GB,  GC  ;  these  arcs  will 
be  equal  to  each  other  (Prop.  V.).  At  the  point  E,  make  the 
angle  DEH  equal  to  the  angle  ABG;  make  the  arc  EH 
equal  to  the  arc  BG  ;  and  join  DH,  FH. 

Because,  in  the  triangles  ABG,  DEH,  the  sides  DE,  EH 
are  equal  to  the  sides  AB,  BG,  and  the  included  angle  DEH 
is  equal  to  ABG ;  the  arc  DH  is  equal  to  AG,  and  the  angle 
DHE  equal  to  AGB  (Prop.  XIII.). 

Now,  because  the  triangles  ABC,  DEF  are  mutually  equi- 
lateral, they  are  mutually  equiangular  (Prop.  XL) ;  hence 
the  angle  ABC  is  equal  to  the  angle  DEF.  Subtracting  the 
equal  angles  ABG,  DEH,  the  remainder  GBC  will  be  equal 
to  the  remainder  HEF.  Moreover,  the  sides  BG,  BC  are 
equal  to  the  sides  EH,  EF ;  hence  the  arc  HF  is  equal  to  the 
arc  GC,  and  the  angle  EHF  to  the  angle  BGC  (Prop.  XIIL). 

Now  the  triangle  DEH.  may  be  applied  to  the  triangle 
ABG  so  as  to  coincide.  For,  place  DH  upon  its  equal  BG 
and  HE  upon  its  equal  AG,  they  will  coincide,  because  the 
angle  DHE  is  equal  to  the  angle  AGB ;  therefore  the  two 
triangles  coincide  throughout,  and  have  equal  surfaces.  For 
the  same  reason,  the  surface  HEF  is  equal  to  the  surface 
GBC,  and  the  surface  DFH  to  the  surface  ACG.  Hence 

ABG+GBC-ACG=DEH+EHF-DFH ; 
or,  ABC  =  DEF; 

that  is,  the  two  triangles  ABC,  DEF  are  equivalent.     There 
fore,  if  two  triangles,  &c. 

Scholium.  The  poles  G  and  H  might  be  situated  within 
the  triangles  ABC,  DEF ;  in  which  case  it  would  be  neces- 
sary to  add  the  three  triangles  ABG,  GBC,  ACG  to  form  the 
triangle  ABC;  and  als3  to  add  the  three  triangles  DEH 


BOOK    IX. 


16J 


EHF,  DFH  to  form  the  triangle  DEF;  otherwise  the  demon- 
stration would  be  the  same  as  above. 

Cor.  If  two  triangles  on  equal  spheres,-are  mutually  equi- 
angular, they  are  equivalent.  They  are  also  equivalent,  if 
they  have  two  sides,  and  the  included  angle  of  the  one,  equal 
to  two  sides  and  the  included  angle  of  the  other,  each  to 
each  ;  or  two  angles  and  the  included  side  of  the  one  equal 
to  two  angles  and  the  included  side  of  the  other 


PROPOSITION   XVI.       THEOREM. 

In  an  isosceles  spherical  triangle,  the  angles  opposite  the 
equal  sides  are  equal;  and,  conversely,  if  two  angles  of  a 
spherical  triangle  are  equal,  the  triangle  is  isosceles. 

Let  ABC  be  a  spherical  triangle,  having 
the  side  AB  equal  to  AC  ;  then  will  the  angle 
ABC  be  equal  to  the  angle  ACB. 

From  the  point  A  draw  the  arc  AD  to  the 
middle  of  the  base  BC.  Then,  in  the  two  tri- 
angles ABD,  ACD,  the  side  AB  is  equal  to 
AC,  BD  is  equal  to  DC,  and  the  side  AD  is 
common  ;  hence  the  angle  ABD  is  equal  to 
the  angle  ACD  (Prop.  XI.). 

Conversely.  Let  the  angle  B  be  equal  to  the 
angle  C  ;  then  will  the  side  AC  be  equal  to 
the  side  AB. 

For  if  the  two  sides  are  not  equal  to  each 
other,  let  AB  be  the  greater ;  take  BE  equal 
to  AC,  and  join  EC.  Then,  in  the  triangles 
EBC,  ACB,  the  two  sides  BE,  BC  are  equal  to 
the  two  sides  CA,  CB,  and  the  included  angles  B 
EBC,  ACB  are  equal ;  hence  the  angle  ECB  is  equal  to  the 
angle  ABC  (Prop.  XIII.).  But,  by  hypothesis,  the  angle  ABC 
_s  equal  to  ACB  ;  hence  ECB  is  equal  to  ACB,  which  is  ab- 
surd. Therefore  AB  is  not  greater  than  AC ;  and,  in  the 
same  manner,  it  can  be  proved  that  it  is  not  less  ;  it  is,  con- 
sequently, equa.  to  AC.  Therefore,  in  an  isosceles  spherical 
triangle,  &c. 

Cor.  The  angle  BAD  is  equal  to  the  angle  CAD,  and  the 
angle  ADB  to  the  angle  ADC  ;  therefore  each  of  the  last  two 
angles  is  a  right  angle.  Hence  the  arc  drawn  from  the  vertex 
of  an  isosceles  spherical  triangle,  to  the  middle  of  the  base  .,# 
perpendicular  to  the  base,  and  bisects  the  vertical  angle. 


If>2 


GEOMETRY 


PROPOSITION   XVII.       THEOREM. 

In  a  sphei  ical  triangle,  the  greater  side  is  opposite  the  greale? 
tngle,  and  conversely. 

Let  ABC  be  a  spherical  triangle,  hav- 
mg  the  angle  A  greater  than  the  angle 
B;  then  will  the  side  BC  be  greater 
than  the  side  AC. 

Diaw  the  arc  AD,  making  the  angle 
BAD  equal  to  B.  Then,  in  the  triangle 
ABD,  we  shall  have  AD  equal  to  DB  B 
(Prop.  XVI.) ;  that  is,  BC  is  equal  to  the  sum  of  AD  and  DC 
But  AD  and  DC  are  together  greater  than  AC  (Prop.  II.)  ; 
hence  BC  is  greater  than  AC. 

Conversely.  If  the  side  BC  is  greater  than  AC,  then  will 
the  angle  A  be  greater  than  the  angle  B.  For  if  the  angle 
A  is  not  greater  than  B,  it  must  be  either  equal  to  it,  or  less. 
It  is  not  equal ;  for  then  the  side  BC  would  be  equal  to  AC 
(Prop.  XVI.),  which  is  contrary  to  the  hypothesis.  Neither 
can  it  be  less ;  for  then  the  side  BC  would  be  less  than  AC, 
by  the  first  case,  which  is  also  contrary  to  the  hypothesis 
Hence  the  angle  BAG  is  greater  than  the  angle  ABC. 
Therefore,  in  a  spherical  triangle,  &c. 


PROPOSITION    XVIII.       THEOREM. 

The  area  of  a  lime  is  to  the  surface  of  the  sphere,  as  the  an- 
gle of  the  lune  is  to  four  right  angles. 

Let  ADBE  be  a  lune,  upon  a  sphere 
whose  center  is  C,  and  the  diameter 
AB ;  then  will  the  area  of  the  lune  be 
to  f  he  surface  of  the  sphere,  as  the  an- 
gle DCE  to  four  right  angles,  or  as  the 
arc  DE  to  the  circumference  of  a  great 
circle. 

First.  When  the  ratio  of  the  arc  to 
the  circumference  can  be  expressed  in 
whole  numbers. 

Suppose  the  ratio  of  DE  to  DEFG  to  be  as  4  to  25.  Now 
if  we  divide  the  circumference  DEFG  in  25  equal  parts,  DE 
will  contain  4  of  those  parts.  If  we  join  the  pole  A  and  the 
several  points  of  division,  bv  arcs  of  great  circles;  there  wilt 


BOOK   IX.  .163 

be  formed  on  the  hemisphere  ADEFG,  25  triangles,  all  equal 
to  each  other,  being  mutually  equilateral.  The  entire  sphere 
will  contain  50  of  these  small  triangles,  and  the  June  ADBE 
8  of  them.  Hence  the  area  of  the  lune  is  to  the  surface  of 
the  sphere,  as  8  to  50,  or  as  4  to  25 ;  that  is,  as  the  arc  DE 
to  the  circumference. 

Secondly.  When  the  ratio  of  the  arc  to  the  circumference 
can  not  be  expressed  in  whole  numbers,  it  may  be  proved,  as 
in  Prop.  XIV.,  B.  III.,  that  the  lune  is  still  to  the  surface  of 
the  sphere,  as  the  angle  of  the  lune  to  four  right  angles. 

Cor.  1.  On  equal  spheres,  two  lunes  are  to  each  other  as 
the  angles  included  between  their  planes. 

Cor.  2.  We  have  seen  that  the  entire  surface  of  the  sphere 
is  equal  to  eight  quadrantal  triangles  (Prop.  X.,  Cor.).  It 
the  area  of  the  quadrantal  triangle  be  represented  by  T,  the 
surface  of  the  sphere  will  be  represented  by  8T.  Also,  if  we 
take  the  right  angle  for  unity,  and  represent  the  angle  of  the 
lune  by  A,  we  shall  have  the  proportion 

area  of  the  lune :  8T  :  :  A  :  4. 

8  A  X  T 

Hence  the  area  of  the  lune  is  equal  to — ,  or  2AxT. 

4 

Cor.  3.  The  spherical  ungula,  comprehended  by  the  planes 
ADB,  AEB,  is  to  the  entire  sphere,  as  the  angle  DCE  is  to 
four  right  angles.  For  the  lunes  being  equal,  the  spherical 
ungulas  will  also  be  equal ;  hence,  in  equal  spheres,  two  un- 
gulas  are  to  each  other  as  the  angles  included  between  their 
planes. 


PROPOSITION    XIX.       THEOREM. 

If  two  great  circles  intersect  each  other  on  the  surface  of  a 
hemispliere,  the  sum  of  the  opposite  triangles  thus  formed,  u 
equivalent  to  a  lune,  whose  angle  is  equal  to  the  inclination  of 

the  two  circles. 

• 

Let  the  great  circles  ABC,  DBE  in- 
tersect each  other  on  the  surface  of 
the  hemisphere  BADGE  ;  then  will  the 
sum  of  the  opposite  triangles  ABD, 
CBE  be  equivalent  to  a  lune  whose 
angle  is  CBE. 

For,  produce  the  arcs  BC,  BE  till 
they  meet  in  F ;  then  will  BCF  be  a 
semicircumference,  as  a'so  ABC.    Sub- 
tracting BC  from  each,  we  shall  have  CF 'equal  to  AB.     For 
the  same  reason  EF  is  equal  to  DB,  and  CE  is  equal  to  AD. 


164  GEOMETRY. 


Hence  the  two  triangles  ABD,  CFE  are  mutua  ly  equilat- 
eral ;  they  are,  therefore,  equivalent  (Prop.  XV.).  But  the 
two  triangles  CBE,  CFE  compose  the  lune  BCFE,  whose  an- 
gle is  CBE ;  hence  the  sum  of  the  triangles  ABD,  CBE  is 
equivalent  to  the  lune  whose  angle  is  CBE.  Therefore,  if 
two  great  circles,  &c. 


PROPOSITION    XX,       THEOREM. 

The  surface  of  a  spherical  triangle  is  measured  by  the  tx 
cess  of  the  sum  of  its  angles  above  two  right  angles,  multiplier 
by  the  quadrantal  triangle. 

Let  ABC  be  any  spherical  triangle ;  its 
surface  is  measured  by  the  sum  of  its  an- 
gles A,  B,  C  diminished  by  two  right  an- 
gles, and  multiplied  by  the  quadrantal  tri- 
angle. 

Produce  the  sides  of  the  triangle  ABC, 
until  they  meet  the  great  circle  DEG, 
drawn  without  the  triangle.  The  two 
triangles  ADE,  AGH  are  together  equal 
to  the  lune  whose  angle  is  A  (Prop.  XIX.)  ;  and  tliits  lune  is 
measured  by  2A  X  T  (Prop.  XVIIL,  Cor.  2).  Henoe  we  have 

ADE+AGH  =  2AxT. 
For  the  same  reason, 

BFG+BDI=2BxT; 
ilso,  CHI+CEF=2CxT. 

But  the  sum  of  these  six  triangles  exceeds  the  surface  of 
the  hemisphere,  by  twice  the  triangle  ABC ;  and  the  hemi- 
sphere is  represented  by  4T  ;  hence  we  have 

4T+2ABC  =  2AxT+2BxT+2CxT; 
or,  dividing  by  2,  and  then   subtracting  2T  from  each  ol 
these  equals,  we  have 

«ABC  =  AxT+BxT-fCxT  —  2T, 

=  (A+B+C-2)XT. 

Hence  every  spherical  triangle  is  measured  by  the  sum  of 
its  angles  diminished  by  two  right  angles,  and  multiplied  by 
the  quadrantal  triangle. 

Cor.  If  the  sum  of  the  three  angles  of  a  triangle  is  equal 
to  three  right  angles,  its  surface  will  be  equal  to  the.  quad- 
rantal triangle ;  if  the  sum  is  equal  to  four  right  angles,  the 
surface  of  the  triangle  will  be  equal  to  two  quadrantal  trian- 
gles ;  if  the  sum  is  equal  to  five  right  angles,  the  surface  will 
be  equal  to  three  quadrantal  triangles,  etc. 


BOOK    IX.  165 


PROPOSITION   XXI.       THEOREM. 

TJie  surface  of  a  spherical  polygon  is  measured  by  the  sum 
(fits  angles,  diminished  by  as  many  times  two  right  anghs  as 
it  has  sides  less  two,  multiplied  by  the  quadrantal  triangle. 

Let  ABODE  be  any  spherical  polygon. 
From  the  vertex  B  draw  the  arcs  BD, 
BE  to.  the  opposite  angles ;  the  polygon 
will  be  divided  into  as  many  triangles  as 
it  has  sides,  minus  two.  But  the  surface 
of  each  triangle  is  measured  by  the  sum 
of  its  angles  minus  two  right  angles,  mul- 
tiplied by  the  quadrantal  triangle.  Also, 
the  sum  of  all  the  angles  of  the  triangles,  is  equal  to  the  sum 
of  all  the  angles  of  the  polygon ;  hence  the  surface  of  the 
polygon  is  measured  by  the  sum  of  its  angles,  diminished  by 
as  many  times  two  right  angles  as  it  has  sides  less  two,  mul- 
tiplied by  the  quadrantal  triangle. 

Cor.  If  the  polygon  has  five  sides,  and  the  sum  of  its  an 
gles  is  equal  to  seven  right  angles,  its  surface  will  be  equal  to 
the  quadrantal  triangle ;  if  the  sum  is  equal  to  eight  right  an- 
gles, its  surface  will  be  equal  to  two  quadrantal  triangles  ;  if 
the  sum  is  equal  to  nine  right  angles,  the  surface  will  be 
equal  to  three  quadrantal  triangles,  etc. 


106  GEOMETRY. 


BOOK     X. 

THE  THREE  ROUND  BODIES. 
Definitions. 

1.  A  cylinder  is  a  solid  described  by  the  revolu- 
tion of  a  rectangle  about  one  of  its  sides,  which 
remains  fixed.     The  bases  of  the  cylinder  are  the 
circles  described  by  the  two  revolving  opposite 
sides  of  the  rectangle. 

2.  The  axis  of  a  cylinder  is  the  fixed  straight 
line  about  which  the  rectangle  revolves.     The  op- 
posite side  of  the  rectangle  describes  the  convex 
surface. 

3.  A  cone  is  a  solid  described  by  the  revolu- 
tion of  a  right-angled  triangle  about  one  of  the 
sides  containing  the  right  angle,  which  side  re- 
mains fixed.     The  base  of  the  cone  is  the  cir- 
cle described  by  that  side  containing  the  right 
angle,  which  revolves. 

4.  The  axis  of  a  cone  is  the  fixed  straight 
line  about  which  the  triangle  revolves.     The 
hypothenuse  of  the  triangle  describes  the  convex  surface. 
The  side  of  the  cone  is  the  distance  from  the  vertex  to  the 
circumference  of  the  base. 

5.  A  frustum  of  a  cone  is  the  part  of  a  cone  next  the 
base,  cut  off  by  a  plane  parallel  to  the  base. 

6.  Similar  cones  and  cylinders  are  those  which  have  their 
axes  and  the  diameters  of  their  bases  proportionals. 


PROPOSITION  I.       THEOREM. 

The  convex  surface  of  a  cylinder  is  equal  to  ike  prcduct  of 
its  altitude  by  the  circumference  of  its  base. 

Let  ACE-G  be  a  cylinder  whose  base  is  the  circle  ACE 
and  altitude  AG ;  then  will  its  convex  surface  be  equal  to 
'he  product  of  AG  by  the  circumference  ACE. 


BOUK    X. 


167 


In  the  circle  ACE  inscribe  the  regular 
polygon  ABCDEF;  and  upon  this  polygon  G 
.et  a  right  prism  be  constructed  of  the  same 
altitude  with  the  cylinder.  The  edges  AG, 
BH,  CK,  &c.,  of  the  prism,  being  perpen- 
dicular to  the  plane  of  the  base,  will  be  con- 
tained in  the  convex  surface  of  the  cylinder. 
The  convex  surface  of  this  prism  is  equal  to 
the  product  of  its  altitude  by  the  perimeter 
of  its  base  (Prop.  L,  B.  VIII.).  Let,  now, 
the  arcs  subtended  by  the  sides  AB,  BC,  &c.,  be  bisected, 
and  the  number  of  sides  of  the  polygon  be  indefinitely  in- 
creased ;  its  perimeter  will  approach  the  circumference  of  the 
circle,  and  will  be  ultimately  equal  to  it  (Prop.  XL,  B.  VI.),- 
and  the  convex  surface  6f  the  prism  will  become  equal  to 
the  convex  surface  of  the  cylinder.  But  whatever  be  the 
number  of  sides  of  .the  prism,  its  convex  surface  is  equal  to 
the  product  of  its  altitude  by  the  perimeter  of  its  base  ;  hence 
the  convex  surface  of  the  cylinder  is  equal  to  the  product  of 
its  altitude  by  the  circumference  of  its  base. 

Cor.  If  A  represent  the  altitude  of  a  cylinder,  and  R  the 
radius  of  its  base,  the  circumference  of  the  base  will  be  repre- 
sented by  2rrR  (Prop.  XIIL,  Cor.  2,  B.  VI.) ;  and  the  convex 
surface  of  the  cylinder  by  2-RA. 


PROPOSITION  II.       THEOREM. 

The  solidity  of  a  cylinder  is  equal  to  the  product  of  its  bast 
by  its  altitude. 

Let  ACE-G  be  a  cylinder  whose  base  is 
the  circle  ACE  and  altitude  AG ;  its  solidity  G 
is  equal  to  the  product  of  its  base  by  its  al- 
titude. 

In  the  circle  ACE  inscribe  the  regular 
polygon  ABCDEF ;  and  upon  this  polygon 
let  a  right  prism  be  constructed  of  the  same 
altitude  with  the  cylinder.  The  solidity  of 
this  prism  is  equal  to  the  product  of  its  base 
by  its  altitude  (Prop.  XL,  B.  VIIL).  Let, 
now,  the  number  of  sides  of  the  polygon  be  indefinitely  in 
creased  ;  its  area  will  become  equal  to  that  of  the  circle,  and 
the  solidity  of  the  prism  becomes  equal  to  that  of  the  cylinder. 
But  whatever  be  the  number  of  sides  of  the  prism,  its  solidity 
is  equal  to  the  product  of  its  base  by  its  altitude ;  hence  the 
solidity  of  a  cylinder  is  equal  to  the  product  of  its  ?ase  by  its 
altitude 


168  GEOMETRY. 

Cor.  1.  If  A  represent  the  altitude  of  a  cylinder,  and  K 
the  radius  of  its  base,  the  area  of  the  base  will  be  represent- 
ed  by  7rRa  (Prop.  XIII.,  Cor.  3,  B.  VI.) ;  and  the  solidity  of 
the  cylinder  will  be  7rR2A. 

Cor.  2.  Cylinders  of  the  same  altitude,  are  to  each  otaer 
as  their  bases;  and  cylinders  of  the  same  base, are  to  each 
other  as  their  altitudes. 

Cor.  3.  Similar  cylinders  are  to  each  other  as  the  cubea 
of  their  altitudes,  or  as  the  cubes  of  the  diameters  of  their 
bases.  For  the  bases  are  as  the  squares  of  their  diameters  ; 
and  since  the  cylinders  are  similar,  the  diameters  of  the  bases 
are  as  their  altitudes  (Def.  6).  Therefore  the  bases  are  as 
the  squares  of  the  altitudes ;  and  hence  the  products  of  the 
bases  by  the  altitudes,  or  the  cylinders  themselves,  will  be  as 
the  cubes  of  the  altitudes. 


PROPOSITION   III.       THEOREM. 

The  convex  surface  of  a  cone  is  equal  to  the  product  of  hall 
its  side,  by  the  circumference  of  its  base. 

Let  A-BCDEFG  be  a  cone  whose  base  is 
the  circle  BDEG,  and  its  side  AB  ;  then  will 
its  convex  surface  be  equal  to  the  product 
of  half  its  side  by  the  circumference  of  the 
circle  BDF. 

In  the  circle  BDF  inscribe  the  regular 
polygon  BCDEFG;  and  upon  this  polygon 
let  a  regular  pyramid  be  constructed  having 
A  for  its  vertex.  The  edges  of  this  pyramid 
will  lie  in  the  convex  surface  of  the  cone. 
From  A  draw  AH  perpendicular  to  CD,  one  of  the  sides  of 
the  polygon.  The  convex  surface  of  the  pyramid  is  equal  to 
the  product  of  half  the  slant  height  AH  by  the  perimeter  of 
its  base  (Prop.  XIV.,  B.  VIII.).  Let,  now,  the  arcs  subtend- 
ed by  the  sides  BC,  CD,  &c.,  be  bisected,  and  the  numbei 
of  sides  of  the  polygon  be  indefinitely  increased,  its  perimeter 
wix.  become  equal  to  the  circumference  of  the  circle,  the  slant 
height  AH  becomes  equal  to  the  side  of  the  cone  AB,  and 
he  convex  surface  of  the  pyramid  becomes  equal  to  the  con- 
vex surface  of  the  cone.  But,  whatever  be  the  number  of 
faces  of  the  pyramid,  its  convex  surface  is  equal  to  the  prod- 
act  of  half  its  slant  height  by  the  perimeter  of  its  base  ;  hence 
the  convex  surface  of  the  cone,  is  equal  to  the  product  of 
half  its  side  by  the  circumference  of  its  base. 

Cor*  If  S  represent  the  side  of  a  cone,  and  R  the  radius 


BOOK    X. 


169 


of  its  base,  then  the  circumference  of  the  base  will  be  repre- 
sented by  27rR,  and   the   convex   surface  of  the   cone  by 


PROPOSITION  IV.       THEOREM. 

The  convex  surface  of  a  frustum  of  a  cone  is  equal  to  th& 
vroduct  of  its  side,  by  half  the  sum  of  the  circumferences  of  its 
two  bases. 

Let  BDF-&e?f  be  a  frustum  of  a  cone 
whose  bases  are  BDF,  bdf,  and  B6  its 
side ;  its  convex  surface  is  equal  to  the 
product  of  Eb  by  half  the  sum  of  the  cir- 
cumferences BDF,  bdf. 

Complete  the  cone  A-BDF  to  which  the 
frustum  belongs,  and  in  the  circle  BDF 
inscribe  the  regular  polygon  BCDEFG ; 
and  upon  this  polygon  let  a  regular  pyr- 
amid be  constructed  having  A  for  its 
vertex.  Then  will  BDF-bdf  be  a  frus- 
tum of  a  regular  pyramid,  whose  convex 
surface  is  equal  to  the  product  of  its  slant  height  by  half  the 
sum  of  the  perimeters  of  its  two  bases  (Prop.  XIV.,  Cor.  1,  B. 
VIII.).  Let,  now,  the  number  of  sides  of  the  polygon  be  in- 
definitely increased,  its  perimeter  will  become  equal  to  the 
circumference  of  the  circle,  and  the  convex  surface  of  the 
pyramid  will  become  equal  to  the  convex  surface  of  the  cone. 
But,  whatever  be  the  number  of  faces  of  the  pyramid,  the  con- 
vex surface  of  its  frustum  is  equal  to  the  product  of  its  slant 
neight,  by  half  the  sum  of  the  perimeters  of  its  two  bases. 
Hence  the  convex  surface  of  a  frustum  of  a  cone  is  equal  to 
the  product  of  its  side  by  half  the  sum  of  the  circumferences 
of  its  two  bases. 

Cor.  It  was  proved  (Prop.  XIV.,  Cor.  2,  B.  VIII.),  that  the 
convex  surface  of  a  frustum  of  a  pyramid  is  equal  to  the 
product  of  its  slant  height,  by  the  perimeter  of  a  section  at 
equal  distances  between  its  two  bases ;  hence  the  convex  sur- 
face of  a  frustum  of  a  cone  is  equal  to  the  product  oj  its  side, 
by  the  circumference  of  a  section  at  equal  distances  between  the 
two  bases 

H 


170  GEOMETRY 


PROPOSITION   V.       THEOREM. 

The  solidity  of  a  cone  is  equal  to  one  third  of  the  inoduct  of 

its  base  and  altitude. 

• 

Let  A-BCDF  be  a  cone  whose  base  is  the 
circle  BCDEFG,  and  AH  its  altitude ;  the 
solidity  of  the  cone  will  be  equal  to  one  thira 
of  the  product  of  the  base  BCDF  by  the  al- 
titude AH. 

In  the  circle  BDF  inscribe  a  regular  poly- 
gon BCDEFG,  and  construct  a  pyramid 
whose  base  is  the  polygon  BDF,  and  having  B( 
its  vertex  in  A.  The  solidity  of  this  pyra- 
mid is  equal  to  one  third  of  the  product  of 
the  polygon  BCDEFG  by  its  altitude  AH  (Prop.  XVIL,  B. 
VIII.).  Let,  now,  the  number  of  sides  of  the  polygon  be  in- 
definitely increased;  its  area  will  become  equal  to  the  area 
of  the  circle,  and  the  solidity  of  the  pyramid  will  become 
equal  to  the  solidity  of  the  cone.  But,  whatever  be  the 
number  of  faces  of  the  pyramid,  its  solidity  is  equal  to  one 
third  of  the  product  of  its  base  and  altitude ;  hence  the  solidity 
of  the  cone  is  equal  to  one  third  of  the  product  of  its  base  and 
altitude. 

Cor.  1.  Since  a  cone  is  one  third  of  a  cylinder  having  the 
*&me  base  and  altitude,  it  follows  that  cones  of  equal  alti 
Judes  are  to  each  other  as  their  bases ;  cones  of  equal  bases 
are  to  each  other  as  their  altitudes ;  and  similar  cones  are  as 
the  cubes  of  their  altitudes,  or  as  the  cubes  of  the  diameters 
of  their  bases. 

Cor.  2.  If  A  represent  the  altitude  of  a  cone,  and  R  the 
radius  of  its  base,  the  solidity  of  the  cone  will  be  represented 
by  7rRa  X  £  A,  or  iTrR'A. 


PROPOSITION    VI.       THEOREM. 

A  frustum  of  a  cone  is  equivalent  to  the  sum  of  three  cones, 
saving  the  same  altitude  with  the  frustum,  and  whose  bases  are 
the  lower  base  of  the  frustum,  its  upper  base,  and  a  mean  pro* 
portional  between  them. 


Let  BDF-fo/f  be  any  frustum  of  a  cone.  Complete  the 
cone  to  which  the  frustum  belongs,  and  in  the  circle  BDF  in 
scribe  the  regular  polygon  BCDEFG ;  and  upon  this  poly 


BOOK    X. 


171 


gon  let  a  regular  pyramid  be  construct- 
ed having  its  vertex  in  A.  Then  will 
BCVEFG-bcdefg  be  a  frustum  of  a  reg- 
ular pyramid,  whose  solidity  is  equal  to 
three  pyramids  having  the  same  altitude 
with  the  frustum,  and  whose  bases  are 
the  lower  base  of  the  frustum,  its  upper 
base,  and  a  mean  proportional  between 
them  (Prop.  XVIIL,  B.  VIII.).  Let,  now, 
the  number  of  sides  of  the  polygon  be  in- 
definitely  increased,  its  area  will  become 
equal  to  the  area  of  the  circle,  and  the 
frustum  of  the  pyramid  will  become  the  frustum  of  a  cone 
Hence  the  frustum  of  a  cone  is  equivalent  to  the  sum  of  three 
cones,  having  the  same  altitude  with  the  frustum,  and  whose 
bases  are  the  lower  base  of  the  frustum,  its  upper  base,  and 
a  mean  proportional  between  them. 


M 


PROPOSITION    VII.       THEOREM. 

The  surface  of  a  sphere  is  equal  to  the  product  of  its  diame 
ter  by  the  circumference  of  a  great  circle. 

Let  ABDF  be  the  semicircle  by  the  revo- 
lution of  which  the  sphere  is  described.  In- 
scribe in  the  semicircle  a  regular  semi-poly- 
gon ABCDEF,  and  from  the  points  B,  C,  D, 
E  let  fall  the  perpendiculars  BG,  CH,  DK, 
EL  upon  the  diameter  AF.  If,  now,  the 
polygon  be  revolved  about  AF,  the  lines  AB, 
EF  will  describe  the  convex  surface  of  two 
cones ;  and  BC,  CD,  DE  will  describe  the 
convex  surface  of  frustums  of  cones. 

From  the  center  I,  draw  IM  perpendicular 
to  BC  ;  also,  draw  MN  perpendicular  to  AF, 
and  BO  perpendicular  to  CH.  Let  circ.  MN  represent  the 
circumference  of  the  circle  described  by  the  revolution  of 
MN.  Then  the  surface  described  by  the  revolution  of  BC, 
will  be  equal  to  BC,  multiplied  by  circ.  MN  (Prop.  IV. 
Cor.). 

Now,  the  triangles  IMN,  BCO  are  similar,  since  their  sides 
are  perpendicular  to  each  other  (Prop.  XXL,  B.  IV.) ;  whence 
BC  :  BO  or  GH  :  :  IM  :  MN, 

:  :  circ.  IM  :  circ.  MN. 
Hence  (Prop.  I.,  B.  II.), 

BC  X  circ.  MN  =  GH  X  circ.  IM. 


172  GEOMETRY. 

Therefore  the  surface  described  by  BC,  is 
equal  to  the  altitude  GH,  multiplied  by  circ.           •"£•-•" 
IM,  or  the  circumference  of  tne  inscribed        /^f 
circle.  //  H 

In  like  manner,  it  may  oe  proved  that  the   c/— — |X: \S 

surface  described  by  CD  is  equal  to  the  alti- 
tude HK,  multiplied  by  the  circumference  of 
he  inscribed  circle;  and  the  same  may  be 
proved  of  the  other  sides.  Hence  the  entire 
surface  described  by  ABCDEF  is  equal  to 
the  circumference  of  the  inscribed  circle,  mul- 
tiplied by  the  sum  of  the  altitudes  AG,  GH, 
HK,  KL,  and  LF ;  that  is,  the  axis  of  the  polygon. 

Let,  now,  the  arcs  AB,  BC,  &c.,  be  bisected,  and  the  num- 
ber of  sides  of  the  polygon  be  indefinitely  increased,  its  pe- 
rimeter will  coincide  with  the  circumference  of  the  semicircle, 
and  the  perpendicular  IM  will  become  equal  to  the  radius  of 
the  sphere  ;  that  is,  the  circumference  of  the  inscribed  circle 
will  become  the  circumference  of  a  great  circle.  Hence  the 
surface  of  a  sphere  is  equal  to  the  product  of  its  diameter  by 
the  circumference  of  a  great  circle. 

Cor.  I.  The  area  of  a  zone  is  equal  to  the  product  of  its  al 
titude  by  the  circumference  of  a  great  circle. 

For  the  surface  described  by  the  lines  BC,  CD  is  equal  to 
the  altitude  GK,  multiplied  by  the  circumference  of  the  in- 
scribed circle.  But  when  the  number  of  sides  of  the  polygon 
is  indefinitely  increased,  the  perimeter  BC-}-CD  becomes  the 
arc  BCD,  and  the  inscribed  circle  becomes  a  great  circle. 
Hence  the  area  of  the  zone  produced  by  the  revolution  of 
BCD,  is  equal  to  the  product  of  its  altitude  GK  by  the  cir 
cumference  of  a  great  circle. 

Cor.  2.  The  area  of  a  great  circle  is  equal  to  the  prod- 
uct of  its  circumference  by  half  the  radius  (Prop.  XII.,  B. 
VI.),  or  one  fourth  of  the  diameter ;  hence  the  surface  of  a 
sphere  is  equivalent  to  four  of  its  great  circles. 

Cor.  3,  The  surface  of  a  sphere  is  equal  to  the  convex  su.r 
face  of  the  circumscribed  cylinder. 

For  the  latter  is  equal  to  the  product  of  its 
altitude  by  the  circumference  of  its  base.  But 
its  base  is  equal  to  a  great  circle  of  the  sphere, 
and  its  altitude  to  the  diameter ;  hence  the 
convex  surface  of  the  cylinder,  is  equal  to  the 
product  of  its  diameter  by  the  circumference 
of  a  great  circle,  which  is  also  the  measure  of 
the  surface  of  a  sphere. 

Cor.  4.  Two  zones  upon  equal  spheres,  are  to  each  othei 
RS  their  altitudes  ;  and  any  zone  is  to  the  surface  of  its 


BOOK    X.  173 

sphere,  as  the  altitude  of  the  zone  is  to  the  diameter  of  the 
sphere. 

Cor.  5.  Let  R  denote  the  radius  of  a  sphere,  D  its  diame- 
ter, C  the  circumference  of  a  great  circle,  and  S  the  surface 
of  the  sphere,  then  we  shall  have 

C  =  2:rR,  or  rrD  (Prop.  XIIL,  Cor.  2,  B.  VI.). 
Also,         S=27rR  x  2R  =  4rrR2,  or  TrD2. 
If  A  represents  the  altitude  of  a  zone,  its  area  will  be 

27rRA. 


PROPOSITION   VIII.       THEOREM. 

The  solidity  of  a  sphere  is  equal  to  one  third  the  product  of 
its  surface  by  the  radius. 

Let  ACEG  be  the  semicircle  by  the  revo- 
lution of  which  the  sphere  is  described.  In- 
scribe in  the  semicircle  a  regular  semi-poly- 
gon ABCDEFG,  and  draw  the  radii  BO, 
CO,  DO,  &c. 

The  solid  described  by  the  revolution  of 
the  polygon  ABCDEFG  about  AG,  is  com- 
posed of  the  solids  formed  by  the  revolu- 
tion of  the  triangles  ABO,  BCO,  CDO,  &c., 
about  AG. 

First.  To  find  the  value  of  the  solid  form- 
ed by  the  revolution  of  the  triangle  ABO. 

From  O  draw  OH  perpendicular  to  AB, 
and  from  B  draw  BK  perpendicular  to  AO. 
The  two  triangles  ABK,  BKO,  in  their  revolution  about  AO, 
will  describe  two  cones  having  a  common  base,  viz.,  the  cir- 
cle whose  radius  is  BK.     Let  area  BK  represent  the  area 
of  the  circle  described  by  the  revolution  of  BK.     Then  the 
solid  described  by  the  triangle  ABO  will  be  represented  by 
Area  BKx^AO  (Prop.  V.). 

Now  the  convex  surface  of  a  cone  is  expressed  by  *n-RS 
(Prop.  III.,  Cor.) ;  and  the  base  of  the  cone  by  7rRa.  Hence 
the  convex  surface  :  base  :  :  TrRS  :  -R2, 

:  :  S  :  R  (Prop.  VIII.,  B.  II.). 

But  AB  describes  the  convex  surface  of  a  cone,  of  which 
BK  describes  the  base  ;  hence 

the  surface  described  by  AB  :  area  BK  :  :  AB    BK 

:  :  AO  :  OH, 
because  the  triangles  ABK,  AHO  are  similar.     Hence 

Area  BKx    AO=   OHx  surface  described  by  AB, 
or     Area  BK  x  |AO  ^  ^OH  x  surface  described  by  AB. 


174  GEOMETRY. 

But  we  have  proved  that  the  solid  de- 
scribed by  the  triangle  ABO,  is  equal  to 
area  BKxjAO;  it  is,  therefore,  equal  to 
iOH  x  surface  described  by  AB. 

Secondly.  To  find  the  value  of  the  solid 
formed  by  the  revolution  of  the  triangle 
BCO. 

Produce  BC  until  it  meets  AG  produced 
in  L.  It  is  evident,  from  the  preceding 
demonstration,  that  the  solid  described  by 
the  triangle  LCO  is  equal  to 

^OM  x  surface  described  by  LC  ; 
and  the  solid  described  by  the  triangle  LBO 
is  equal  to 

iOM  X  surface  described  by  LB  ; 

hence  the  solid  described  by  the  triangle  BCO  is  equal  to 
^OM  X  surface  described  by  BC. 

In  the  same  manner,  it  may  be  proved  that  the  solid  de- 
scribed by  the  triangle  CDO  is  equal  to 

iON  x  surface  described  by  CD  ; 

and  so  on  for  the  other  triangles.  But  the  perpendiculars 
OH,  OM,  ON,  &c.,  are  all  equal ;  hence  the  solid  described 
by  the  polygon  ABCDEFG,  is  equal  to  the  surface  described 
by  the  perimeter  of  the  polygon,  multiplied  by  -|OH. 

Let,  now,  the  number  of  sides  of  the  polygon  be  indefinite- 
ly increased,  the  perpendicular  OH  will  become  the  radius 
OA,  the  perimeter  ACEG  will  become  the  semi-circumference 
ADG,  and  the  solid  described  by  the  polygon  becomes  a 
sphere ;  hence  the  solidity  of  a  sphere  is  equal  to  one  third 
jf  the  product  of  its  surface  by  the  radius. 

Cor.  1.  The  solidity  of  a  spherical  sector  is  equal  to  the  prod- 
uct of  the  zone  which  forms  its  base,  by  one  third  of  its  radius. 

For  the  solid  described  by  the  revolution  of  BCDO  \» 
equal  to  the  surface  described  by  BC+CD,  multiplied  b; 
iOM.  But  when  the  number  of  sides  of  the  polygon  is  in 
definitely  increased,  the  perpendicular  OM  becomes  the 
radius  OB,  the  quadrilateral  BCDO  becomes  the  sector 
BDO,  and  the  solid  described  by  the  revolution  of  BCDO 
becomes  a  spherical  sector.  Hence  the  solidity  of  a  spheri- 
cal sector  is  equal  to  the  product  of  the  zone  which  forms  its 
base,  by  one  third  of  its  radius. 

Cor.  2.  Let  R  represent  the  radius  of  a  sphere,  D  its  di- 
ameter, S  its  surface,  and  V  its  solidity,  then  we  shall  have 

S=47rR2  or  TrD3  (Prop.  VII.,  Cor.  5). 
Also,  V=^RxS  =  |7rR3  or  |7rD3; 

hence  the  solidities  of  svheres  are  to  each  other  as  the  cubes  of 
their  radii 


BOOK   X. 


If  we  put  A  to  represent  the  altitude  of  the  zone  which 
forms  the  base  of  a  sector,  then  the  solidity  of  the  sector  will 
be  represented  by 


. 

Cor.  3.  Every  sphere  is  two  thirds  of  the  circumscribed 
cylinder. 

For,  since  the  base  of  the  circumscribed  cylinder  is  equal 
lo  a  great  circle,  and  its  altitude  to  a  diameter,  the  solidity 
of  the  cy.inder  is  equal  to  a  great  circle,  multiplied  by  the 
diameter  (Prop-  II.).  But  the  solidity  of  a  sphere  is  equal 
to  four  great  circles,  multiplied  by  one  third  of  the  radius  ;  or 
one  great  circle,  multiplied  by  f  of  the  radius,  or  f  of  the 
diameter.  Hence  a  sphere  is  two  thirds  of  the  circumscribed 
cylinder. 


PROPOSITION    IX.       THEOREM. 

A  spherical  segment  with  one  base,  is  equivalent  to  half  of 
i  cylinder  having  the  same  base  and  altitude,  plus  a  sphere 
whose  diameter  is  the  altitude  of  the  segment. 

Let  BD  be  the  radius  of  the  base  of  the 
segment,  AD  its  altitude,  and  let  the  segment 
be  generated  by  the  revolution  of  the  circu- 
lar half  segment  AEBD  about  the  axis  AC. 
Join  CB,  and  from  the  center  C  draw  CF  per- 
pendicular  to  AB. 

The  solid  generated  by  the  revolution  of 
the  segment  AEB,  is  equal  to  the  difference  of  the  solids  gen- 
erated by  the  sector  ACBE,  and  the  triangle  ACB.  Now, 
the  solid  generated  by  the  sector  ACBE  is  equal  to 

|7rCB3xAD  (Prop.  VIIL,  Cor.  2). 

And  the  solid  generated  by  the  triangle  ACB,  by  Prop.  VIIL, 
is  equal  to  ^CF,  multiplied  by  the  convex  surface  described 
by  AB,  which  is  SrrCF  x  AD  (Prop.  VII.),  making  for  the  "solid 
generated  by  the  triangle  ACB, 

|TrCF2xAD. 

Therefore  the  solid  generated  by  the  segment  AEB, is  equal 
to  |7rADx(CB2— CF2), 

or  fTrADxBF2; 

that  is,  iTrADxAB2,  " 

because  CB2— CF2  is  equal  to  BF2,  and  BFa  is  equal  to  one 
fourth  of  AB2. 

Now  the  cone  generated  by  the  triangle  ABD  is  equal  to 

iTiADxBD2  (Prop.  V.,  Cor.  2). 

Therefore  the  spherical  segment  in  question,  which  is  the 
sum  of  the  solids  described  by  AEB  and  ABD,  is  equal  to 


176 


GEOMETRY 


that  is,  i7rAD(3BD2-f  AD2), 

because  ABa  is  equal  to  BD2  +  AD2. 

This  expression  may  be  separated  into  the  two  parts 


and 

The  first  part  represents  the  solid- 
ity of  a  cylinder  having  the  same 
base  with  the  segment  and  half  its 
altitude  (Prop.  II.)  ;  the  other  part 
represents  a  sphere,  of  which  AD  is 
the  diameter  (Prop.  VIII.,  Cor.  2). 
segment,  &c. 

Cor.  The  solidity  of  the  spherical  seg- 
ment of  two  bases,  generated  by  the  revolu- 
tion of  BCDE  about  the  axis  AD,  may  be 
found  by  subtracting  that  of  the  segment  of 
one  base  generated  by  ABE,  from  that  of  the 
segment  of  one  base  generated  by  ACD. 


Therefore,  a  spherical 


7 


j\ 


D 


CONIC  SECTIONS, 


THERE  are  three  curves  whose  properties  are  extensively 
applied  in  Astronomy,  and  many  other  brunches  of  science, 
which,  being  the  sections  of  a  cone  made  by  a  plane  in  dif 
ferent  positions,  are  called  the  conic  sections.  These  are 

The  Parabola, 

The  Ellipse,  and 

The  Hyperbola. 


PARABOLA. 

Definitions. 

1.  A  parabola  is  a  plane  curve,  every  point  of  which  is 
equally  distant  from  a  fixed  point,  and  a  given  straight  line. 

2.  The  fixed  point  is  called  the  focus  of  the  parabola  and 
the  given  straight  line  is  called  the  directrix. 

Thus,  if  F  be  a  fixed  point,  and  BC  a   B 
given  line,  and  the  point  A  move  about  F  A    ^ 

in  such  a  manner,  that  its  distance  from  F   D 
is  always  equal  to  the  perpendicular  dis- 
tance from  BC,  the  point  A  will  describe 
a  parabola,  of  which  F  is  the  focus,  and 
BC  the  directrix. 

3.  A  diameter  is  a  straight  line  drawn 
through  any  point  of  the  curve   perpen- 
dicular to  the  directrix.      The  vertex  of 
the  diameter  is  the  point  in  which  it  cuts   c 
the  curve. 

Thus,  through  any  point  of  the  curve,  as  A,  draw  a  line 
DE  perpendicular  to  the  directrix  BC ;  DE  is  a  diameter  oi 
the  parabola,  and  the  point  A  is  the  vertex  of  this  diameter. 

4.  The  axis  of  the  parabola  is  the  diameter  which  passes 
through  the  focus ;  and  the  point  in  which  it  cuts  the  curve 
is  called  the  principal  vertex. 

Thus,  draw  a  diameter  of  the  parabola,  GH,  through  the 


178 


CONIC    SECTIONS. 


focus  F  ;  GH  is  the  axis  of  the,  parabola,  B 
and  the  point  V,  where  the  axis  cuts  the  ^ 
curve,  is  called  the  principal  vertex  of 
the  parabola,  or  simply  the  vertex. 

It  is  evident  from  Def.  1,  that  the  line  3 
FH  is  bisected  in  the  point  V. 

5.  A  tangent  is  a  straight  line  which  E 
meets  the  curve,  but,  being  produced,  does 
not  cut  it.  C 

6.  An  ordinate  to  a  diameter,  is  a  straight  line  drawn  from 
any  point  of  the  curve  to  meet  that  diameter,  and  is  parallel 
to  the  tangent  at  its  vertex. 

Thus,  let  AC  be  a  tangent  to  the 
parabola  at.  B,  the  vertex  of  the  di- 
ameter BD.  From  any  point  E  of  the 
curve,  draw  EGH  parallel  to  AC  ; 
then  is  EG  an  ordinate  to  the  diame- 
ter BD. 

It  is  proved  in  Prop.  IX.,  that  EG 
is  equal  to  GH  ;  hence  the  entire  line 
EH  is  called  a  double  ordinate. 

7.  An  abscissa  is  the  part  of  a  diameter  intercepted  be- 
tween its  vertex  and  an  ordinate. 

Thus,  BG  is  the  abscissa  of  the  diameter  BD,  correspond- 
ing to  the  ordinate  EG. 

8.  A  subtangent  is  that  part  of  a  diameter  intercepted  be- 
tween a  tangent  and  ordinate  to  the  point  of  contact. 

Thus,  let  EL,  a  tangent  to  the  curve  at  E,  meet  the  di- 
ameter BD  in  the  point  L  ;  then  LG  is  the  subtangent  of  BD, 
corresponding  to  the  point  E. 

9.  The  parameter  of  a  diameter  is  the  double  ordinate 
which  passes  through  the  focus. 

Thus,  through  the  focus  F,  draw  IK  parallel  to  the  tan- 
gent AC ;  then  is  IK  the  parameter  of  the  diameter  BD. 

10.  The  parameter  of  the  axis  is  called  the  principal  pa- 
rameter, or  latus  rectum. 

11.  A  normal  is  a  line  drawn  perpendicular  to  a  tangent 
from  the  point  of  contact,  and  terminated  by  the  axis. 

12.  A.  subnormal  is  the  part  of  the  axis 
intercepted  between  the  normal,  and  the 
corresponding  ordinate. 

Thus,  let  AB  be  a  tangent  to  the 
parabola  at  any  point  A.  From  A 
draw  AC  perpendicular  to  AB ;  draw, 
also,  the  ordinate  AD.  Then  AC  is  the 
normal,  and  DC  is  the  subnormal  cor- 
responding to  the  point  A 


PARABOLA. 


PROPOSITION   I.       PROBLEM. 

To  describe  a  parabola. 

Let  BC  be  a  ruler  laid  upon  a  plane, 
and  lei  DEG  be  a  square.  Take  a 
thread  equal  in  length  to  EG,  and  attach 
one  extremity  at  G,  and  the  other  at 
some  point  as  F.  Then  slide  the  side 
of  the  square  DE  along  the  ruler  BC, 
and,  at  the  same  time,  keep  the  thread 
continually  tight  by  means  of  the  pencil 
A ;  the  pencil  will  describe  one  part  of 
a  parabola,  of  which  F  is  the  focus,  and 
BC  the  directrix.  For,  in  every  posi- 
tion of  the  square, 

AF+AG=AE+AG, 
and  hence  AF= AE  ; 

that  is,  the  point  A  is  always  equally  distant  from  the  focus 
F  and  directrix  BC. 

If  the  square  be  turned  over,  and  moved  on  the  other  side 
of  the  point  F,  the  other  part  of  the  same  parabola  may  be 
iescribed. 


PROPOSITION    II.       THEOREM. 

A  tangent  to  the  parabola  bisects  the  angle  formed  at  the 
point  of  contact,  by  a  perpendicular  to  the  directrix,  and  a  lin* 
drawn  to  the  focus. 

Let  A  be  any  point  of  the  parabola 
from  which  draw  the  line  AF  to 
the  focus,  and  AB  perpendicular  to  the 
directrix,  and  draw  AC  bisecting  the  an- 
gle BAF ;  then  will  AC  be  a  tangent  to 
the  curve  at  the  point  A. 

For,  if  possible,  let  the  line  AC  meet 
the  curve  in  some  other  point  as  D. 
Join  DF,  DB,  and  BF ;  also,  draw  DE 
perpendicular  to  the  directrix. 

Since,  in  the  two  triangles  ACB,  ACF,  AF  is  equal  to  AB 
(Def.  1),  AC  is  common  to  both  triangles,  and  the  angle  CAB 
is,  by  supposition,  equal  to  the  angle  CAF ;  therefore  CB  is 
equal  to  CF,  and  the  angle  ACB  to  the  angle  ACF. 


180  CONIC    SECTIONS. 

Again,  in  the  two  triangles  DCB,  DCF,  because  B(J  ,s 
equal  to  CF,  the  side  DC  is  common  to  both  triangles,  and 
the  angle  DCB  is  equal  to  the  angle  DCF ;  therefore  DB  is 
equal  to  DF.  But  DF  is  equal  to  DE  (Def.  1) ;  hence  DB 
is  equal  to  DE,  which  is  impossible  (Prop.  XVII.,  B.  L). 
Therefore  the  line  AC  does  not  meet  the  curve  in  D  ;  and  in 
the  same  manner  it  may  be  proved  that  it  does  not  meet  the 
curve  in  any  other  point  than  A ;  consequently  it  is  a  tangent 
to  the  parabola.  Therefore,  a  tangent,  &c. 

Cor.  1.  Since  the  angle  FAB  continually,  increases  as  the 
point  A  moves  toward  V,  and  at  V  becomes  equal  to  two 
right  angles,  the  tangent  at  the  principal  vertex  is  perpendicu- 
lar to  the  axis.  The  tangent  at  the  vertex  V  is  called  the. 
vertical  tangent. 

Cor.  2.  Since  an  ordinate  to  any  diameter  is  parallel  to 
the  tangent  at  its  vertex,  an  ordinate  to  the  axis  is  perpen 
dicular  to  the  axis. 


PROPOSITION  III.       THEOREM. 

The  lotus  rectum  is  equal  to  four  times  the  distance  from  tht 
focus  to  the  vertex. 

Let  AVB  be  a  parabola,  of  which  F  is  the 
focus,  and  V  the  principal  vertex ;  then  the   ~ 
latus   rectum   AFB   will   be   equal   to   four 
times  FV. 

Let  CD  be  the  directrix,  and  let  AC  be 
drawn  perpendicular  to  it;  then,  according  D 
to  Def.  1,  AF  is  equal  to  AC  or  DF,  because 
ACDF  is  a  parallelogram.  But  DV  is  equal 
to  VF ;  that  is,  DF  is  equal  "to  twice  VF. 
Hence  AF  is  equal  to  twice  VF.  In  the 
same  manner  it  may  be  proved  that  BF  is  equal  to  twice 
VF ;  consequently  AB  is  equal  to  four  times  VF.  There- 
fore, the  latus  rectum,  &c. 


PROPOSITION  IV.       THEOREM. 

If  a  tangent  to  the  parabola  cut  the  axis  produced,  the 
points  of  contact  and  of  intersection  are  equally  distant  from 
the  focus. 

Let  AB  be  a  tangent  to  the  parabola  GAH  at  the  point  A, 
and  let  it  cut  the  axis  produced  in  B ;  also,  let  AF  be  drawn 
to  the  focus  ;  then  will  the  line  AF  be  equal  tc  BF. 


PARABOLA. 


181 


Draw  AC  perpendicular  to  the  di- 
rectrix; then,  since  AC  is  parallel  to 
BF,  the  angle  BAG  is  equal  to  ABF. 
But  the  angle  BAG  is  equal  to  BAF 
(Prop.  II.) ;  hence  the  angle  ABF  Is 
equal  to  BAF,  and,  consequently,  AF 
is  equal  to  BF.  Therefore,  if  a  tan- 
gent, &c. 

Cor.  1.  Let  the  normal  AD  be 
drawn.  Then,  because  BAD  is  a 
right  angle,  it  is  equal  to  the  sum  of  the  two  angles  ABD 
ADB,  or  to  the  sum  of  the  two  angles  BAF,  ADB.  Take 
away  the  common  angle  BAF,  and  we  have  the  angle  DAF 
equal  to  ADF.  Hence  the  line  AF  is  equal  to  FD.  There- 
fore, if  a  circle  be  described  with  the  center  F,  and  radius  FA, 
it  will  pass  through  the  three  points  B,  A,  D. 

Cor.  2.  The  normal  bisects  the  angle  made  by  the  diameter 
at  the  point  of  contact,  with  the  line  drawn  from  that  point  to 
the  focus. 

For,  because  BD  is  parallel  to  CE,  the  alternate  angles 
ADF,  DAE  are  equal.  But  the  angle  ADF  has  been  proved 
equal  to  DAF;  hence  the  angles  DAF,  DAE  are  equal  to 
each  other. 

Scholium.  It  is  a  law  in  Optics,  that  the  angle  made  by  a 
ray  of  reflected  light  with  a  perpendicular  to  the  reflecting 
surface,  is  equal  to  the  angle  which  the  incident  ray  makes 
with  the  same  perpendicular.  Hence,  if  GAH  represent  a 
concave  parabolic  mirror,  a  ray  of  light  falling  upon  it  in  the 
direction  EA  would  be  reflected  to  F.  The  same  would  be 
true  of  all  rays  parallel  to  the  axis.  Hence  the  point  F,  in 
which  all  the  rays  would  intersect  each  other,  is  called  the 
fccus,  or  burning  point. 


PROPOSITION    V.       THEOREM. 

The  subtangent  to  the  axis  is  bisected  by  the  vertex. 


Let  AB  be  a  tangent  to  the  parab- 
o  a  ADV  at  the  point  A,  and  AC 
an  ordinate  to  the  axis;  then  wi.l 
BC  be  the  subtangent,  and  it  will  be 
bisected  at  the  vertex  V. 

For  BF  is  equal  to  AF  (Prop. 
IV.) ;  and  AF  is  equal  to  CE,  which 
is  the  distance  of  the  point  \.  from 
the  directrix.  But  CE  is  e^ual  to 
the  sum  of  CV  and  VE,  or  (TV  and 


VF.     Hence  BF,  01 


182 


CONIC   SECTIONS. 


BV+VF  is  jqual   .o  CV+VF ;  that  is,  BV  is  equal  to  C  V 
Therefore,  the  subtangent,  &c. 

Cor.  1.  Hence  the  tangent  at  D,  the  extremity  of  the  latus 
~ectum,  meets  the  axis  in  E,  the  same  point  with  the  direc- 
trix. For,  by  Def.  8,  Ef1  is  the  subtangent  corresponding  to 
the  tangent  DE. 

Cor.  2.  Hence,  if  it  is  required  to  draw  a  tangent  to  the 
curve  at  a  given  point  A,  draw  the  ordinate  AC  to  the  axis, 
Make  BV  equal  to  VC ;  join  the  points  B,  A,  and  the  line 
BA  will  be  the  tangent  required. 


PROPOSITION   VI.       THEOREM. 

The  subnormal  is  equal  to  half  the  latus  rectum. 

Let  AB  be  a  tangent  to  the  parab- 
ola AV  at  the  point  A,  let  AC  be 
he  ordinate,  and  AD  the  normal  from 
the  point  of  contact ;  then  CD  is  the 
subnormal,  and  is  equal  to  half  the  • 
latus  rectum. 

For  the  distance  of  the  point  A 
from  the  focus,  is  equal  to  its  distance 
from  the  directrix,  which  is  equal  to 
VF+VC,  or  2VF+ FC ;  that  is, 

FA=2VF+FC, 

or  2VF=FA~FC. 

Also,  CD  is  equal  to  FD— FC,  which  is  equal  to  FA— FC 
(Prop.  IV.,  Cor.  1).  Hence  CD  is  equal  to  2VF,  which  is 
equal  to  half  the  latus  rectum  (Prop.  III.).  Therefore,  the 
subnormal,  &c. 


PROPOSITION   VII.       THEOREM. 

If  a  perpendicular  be  drawn  from  the  focus  to  any  tangent, 
the  point  of  intersection  will  be  in  the  vertical  tangent. 

Let  AB  be  any  tangent  to  the  pa- 
rabola AV,  and  FC  a  perpendicular  let 
fall  from  the  focus  upon  AB ;  join  VC  ; 
then  will  the  line  VC  be  a  tangent  to 
the  curve  at  the  vertex  V. 

Draw  the  ordinate  AD  to  the  axis 
Since  FA  is  equal  to  FB  (Prop.  IV.), 
and  FC  is  drawn  perpendicular  to 
A.B.  it  divide?  the  triangle  AFB  into 


PARABOLA.  183 

two  equal  parts,  and,  therefore,  AC  is  equal  to  BC.  Bu* 
BV  is  equa.  to  VD  (Prop.  V.) ;  hence 

BC  :  CA  :  :  BV  :  VD, 

and,  therefore,  C  V  is  parallel  to  AD  (Prop.  XVL,  B.  IV.).    But 
AD  is  perpendicular  to  the  axis  BD ;  hence  CV  is  also  per 
pendicular  to  the  axis,  and  is  a  tangent  to  the  curve  at  the 
point  V  (Prop.  II.,  Cor.  1).     Therefore,  if  a  perpendicular, 
&c. 

Cor.  1.  Because  the  triangles  FVC,  FCA  are  similar,  we 
have  FV  :  FC  :  :  FC  :  FA ; 

that  is,  the  perpendicular  from  the  focus  upon  any  tangent,  is  a 
mean  proportional  between  the  distances  of  the  focus  from  the 
vertex,  and  from  the  point  of  contact. 

Cor.  2.  If  is  obvious  that  F  V:  FA : :  FC2:  FA2.(Prop.  XII.,  B.  II.) 

Cor.  3.  From  Cor.  1,  we  have 

FC2=FVxFA. 

But  FV  remains  constant  for  the  same  parabola ;  therefore 
the  distance  from  the  focus  to  the  point  of  contact,  varies  as  the 
square  of  the  perpendicular  upon  the  tangent. 


PROPOSITION    VIII.       THEOREM. 

The  square  of  an  ordinate  to  the  axis,  is  equal  to  the  product 
of  the  lotus  rectum  by  the  corresponding  abscissa. 

Let  AVC  be  a  parabola,  and  A  any  point 
of  the  curve.  From  A  draw  the  ordinate 
AB ;  then  is  the  square  of  AB  equal  to  the 
product  of  VB  by  the  latus  rectum. 

For        AB3  is  equal  to  AF2— FB2. 
But  AF  is  equal   to  VB+VF,  and  FB  is 
equal  to  VB-VF. 

Hence     AB2=(VB-hVF)2-  (VB-VF)2, 
which,  according  to  Prop.  IX.,  Cor.,  B.  IV., 
is  equal  to 

4VBXVF, 

or  VBx  the  latus  rectum  (Prop.  III.). 

Therefore,  the  square,  &c. 

Cor.  1.  Since  the  latus  rectum  is  constant  for  the  same 
parabola,  the  squares  of  ordinates  to  the  axis,  are  to  each  other 
a?  their  corresponding  abscissas. 

Cor.  2.  The  preceding  demonstration  is  equally  applicable 
to  ordinates  on  either  side  of  the  axis ;  hence  AB  is  equal  to 
BC,  and  AC  is  called  a  double  ordinate.  The  curve  is  sym- 
metrical with  respect  to  the  axis,  and  the  wh^lo  pr-.-.ibola  is 
bisected  by  the  axis. 


184 


CONIC  SECTIONS. 


PROPOSITION   IX.       THEOREM. 


The  square  of  an  ordinate  to  any  diameter,  is  equal  to  foui 
times  the  product  of  the  corresponding  abscissa,  by  the  distance 
from  the  vertex  of  that  diameter  to  the  focus. 

Let  AD  be  a  tangent  to  the 
parabola  VAM  at  the  point 
A;  through  A  draw  the  di- 
ameter HAG,  and  through 
any  point  of  the  curve,  as  B, 
draw  BC  parallel  to  AD ; 
draw  also  AF  to  the  focus ; 
then  will  the  square  of  BC  be 
equal  to4AFxAC. 

Draw  CE  parallel,  and  EBG 
perpendicular  to  the  directrix  HK ;  and  join  BH,  BF,  HF, 
Also,  produce  CB  to  meet  HF  in  L. 

Because  the  right-angled  triangles  FHK,  HCL  are  simi- 
lar, and  AD  is  parallel  to  CL,  we  have 

HF : FK : : HC : HL 
:  :  AC  :  DL. 
Hence  (Prop.  L,  B.  II.), 

HFxDL=  FKxAC, 

or  2HF  X  DL  = 2FK  X  AC,  or  4VF  X  AC. 

But  2HFxDL=HL2-LF2  (Prop.  X.,  B.  IV.) 

=  HB2— BF2 
•  =HG2orCE3. 

Hence  CE2  is  equal  to  4VFxAC. 
Also,  because  the  triangles  BCE,  AFD  are  similar,  we  have 


CE 

Therefore  CE2 


CB 
CB2 


DF  :  AF. 

DF  :  AF2  (Prop.  X.,  B.  II.) 

VF  :  AF  (Prop.  VII.,  Cor.  2) 

4VFxAC:4AFxAC. 

But  the  two  antecedents  of  this  proportion  have  been  prove/I 
to  be  equal ;  hence  the  consequents  are  equal,  or 

BC2=4AFxAC. 
Therefore,  tne  square  of  an  ordinate,  &c. 

Cor.  In  like  manner  it  may  be  proved  that  the  square  of 
CM  is  equal  to  4AF  x  AC.  Hence  BC  is  equal  to  CM  ;  raid 
since  the  same  may  be  proved  for  any  ordinate,  it  follows 
that  every  diameter  bisects  its  double  ordinates. 


PARABOLA. 


185 


PROPOSITION    X.       THEOREM.. 

The  parameter  of  any  diameter,  is  equal  to  four  times 
distance  from  its  vertex  to  the  focus. 

•  Let  BAD  be  a  parabola,  of  which 
F  is  the  focus,  AC  is  any  diameter, 
and  BD  its  parameter;  then  is  BD 
equal  to  four  times  AF. 

Draw  the  tangent  AE ;  then,  since 
AEFC  is  a  parallelogram,  AC  is  equal 
to  EF,  which  is  equal  to  AF  (Prop. 
IV.). 

Now,  by  Prop.  IX.,  BC2  is  equal  to 
4AF  X  AC  ;  that  is,  to  4AF2.     Hence 
BC  is  equal  to  twice  AF,  and  BD  is  equal  to  four  times  AF 
Therefore,  the  parameter  of  any  diameter,  &c. 

Cor.  Hence  the  square  of  an  ordinate  to  a  diameter,  is 
equal  to  the  product  of  its  parameter  by  the  corresponding 
abscissa. 


PROPOSITION   XI.       THEOREM. 


If  a  cone  be  cut  by  a  plane  parallel  to  its  side,  the  section  ts 
a  parabola. 

Let  ABGCD  be  a  cone  cut  by  a  plane 
VDG  parallel  to  the  slant  side  AB ;  then 
will  the  section  DVG  be  a  parabola. 

Let  ABC  be  a  plane  section  through 
the  axis  of  the  cone,  and  perpendicular  to 
the  plane  VDG ;  then  VE,  which  is  their 
common  section,  will  be  parallel  to  AB. 
Let  bgcd  be  a  plane  parallel  to  the  base 
of  the  cone ;  the  intersection  of  this  plane 
with  the  cone  will  be  a  circle.  Since  the  B 
plane  ABC  divides  the  cone  into  two 
equal  parts,  BC  is  a  diameter  of  the  circle  G- 

BGCD,  and  be  is  a  diameter  of  the  circle  bgcd.  Let  DEG 
deg  be  the  common  sections  of  the  plane  VDG  with  the 
planes  BGCD,  bgcd  respectively.  Then  DG  is  perpendicular 
to  the  plane  ABC,  and,  consequently,  to  the  lines  VE,  BC. 
For  the  same  reason,  dg  is  perpendicular  to  the  two  lines 
VE,  be. 


186 


CONIC    SECTIONS. 


Now,  since  be  is  parallel  to  BE,  ana 
bE  to  eE,  the  figure  6BEe  is  a  parallelo- 
gram, and  be  is  equal  to  BE.  But  be- 
cause the  triangles  Vec,  VEC  are  similar, 
we  have 

ec  :  EC  :  :  Ve  :  VE  ; 

and  multiplying  the  first  and  second  terms 
of  this  proportion  by  the  equals  be  and 
BE,  we  have 

foXeciBExEC::  Ve:  VE. 
But  since  be  is  a  diameter  of  the  circle 
bgcd,  and  de  is  perpendicular  to  be  (Prop. 
XXII.,  Cor.,  B.  IV.), 

bey.ec   =  de\ 

For  the  same  reason,  BExEC=DEa. 

Substituting  these  values  of  beXec  and  BExEC  in  the  pre- 
ceding proportion,  we  have 

de9  :  DE3  :  :  Ve  :  VE ; 

that  is,  the  squares  of  the  ordinates  are  to  each  other  as  the 
corresponding  abscissas ;  and  hence  the  curve  is  a  parabola, 
whose  axis  is  VE  (Prop.  VIIL,  Cor.  1.).  Hence  the  .-nrab- 
ola  is  called  a  conic  section,  as  mentioned  on  page  177. 


PROPOSITION    XII.       THEOREM. 

Every  segment  of  a  parabola  is  two  thirds  of  its  circum 
scribing  rectangle. 

Let  AVD  be  a  segment  of 
a  parabola  cut  off  by  the 
straight  line  AD  perpendicu- 
lar to  the  axis ;  the  area  of 
AVD  is  two  thirds  of  the  cir- 
cumscribing rectangle  ABCD. 

Draw  the  line  AE  touching 
the  parabola  at  A,  and  meet- 
ing the  axis  produced  in  E; 
and  take  a  point  H  in  the 
surve,  so  near  to  A  that  the 
tangent  and  curve  may  be  regarded  as  eoincicj^ng.  Through 
H  draw  KL  perpendicular,  and  MN  parallel  to  the  axis, 
'"'hen  the 

rectangle  AL  :  rectangle  AM  :  :  AGxGL  :  ABx  AN 

::  AGxGE:  ABxAG 
:  :  GE     A  B, 


i) 


PARABOLA.  187 

because  GL  or  NH  :  AN  :  :  GE  :  AG.  But  GE  is  equal  to 
twice  GV  or  AB  (Prop.  V.) ;  hence 

AL : AM  :  : 2 :  1 ; 

that  is,  AL  is  double  of  AM. 

Hence  the  portion  of  the  parabola  included  between  two  or- 
dinates  indefinitely  near,  is  double  the  corresponding  portion 
of  the  external  space  ABV.  Therefore,  since  the  same  is 
true  for  every  point  of  the  curve,  the  whole  space  AVG  is 
double  the  space  ABV.  Whence  AVG  is  two  thirds  of 
ABVG;  and  the  segment  AVD  is  two  thirds  of  the  rectai*- 
gle  ABCD.  Therefore,  every  segment,  &c 


188 


•;ONIO    SECTIONS. 


ELLIPSE. 

Definitions. 

1 .  AN  ellipse  is  a  plane  curve,  in  which  the  sum  of  the  dis- 
tances of  each  point  from  two  fixed  points,  is  equal  to  a  given 
line. 

2.  The  two  fixed  points  are  called  the  foci. 
Thus,  if  F,  F'  are  two  fixed  points, 

and  if  the  point  D  moves  about  F  in 
such  a  manner  that  the  sum  of  its  dis- 
tances from  F  and  F'  is  always  the 
same,  the  point  D  will  describe  an 
ellipse,  of  which  F  and  F'  are  the  foci. 

3.  The  center  is  the  middle  point  of 
the  straight  line  joining  the  foci. 

4.  The  eccentricity  is  the  distance  from  the  center  to  either 
focus. 

Thus,  let  ABA'B'  be  an  ellipse,  B 

F  and  F'  the  foci.     Draw  the  line 
FF'  and  bisect  it  in  C.     The  point 
C  is  the  center  of  the  ellipse  ;  and     , 
CF  or  CF'  is  the  eccentricity. 

5.  A  diameter  is  a  straight  line 
drawn    through   the    center,   and 
terminated    both    ways    by    the 
curve. 

6.  The  extremities  of  a  diameter  are  called  its  vertices. 
Thus,  through  C  draw  any  straight  line  DD'  terminated 

by  the  curve ;  DD'  is  a  diameter  of  the  ellipse ;  D  and  D' 
are  its  vertices. 

7.  The  major  axis  is  the  diameter  which  passes  through 
the  foci ;  and  its  extremities  are  called  the  principal  vertices. 

S.  The  minor  axis  is  the  diameter  which  is  perpendicular 
to  the  major  axis. 

Thus,  produce  the  line  FF'  to  meet  the  curve  in  A  and 
A' ;  and  through  C  draw  BB'  perpendicular  to  AA' ;  then  is 
AA'  the  major  axis,  and  BB'  the  minor  axis. 

9.  A  tangent  is  a  straight  line  which  meets  the  curve,  but 
Deing  produced,  does  not  cut  it. 


D' 


ELLIPSE. 


189 


(r 


10.  An  ordinate  to  a  diameter,  is  a  straight  line  drawn 
from  any  point  of  the  crave  to  the  diameter,  parallel  to  the 
tangent  at  one  of  its  vertices. 

Thus,  let  DD'  be  any  diameter, 
and  TT'  a  tangent  to  the  ellipse 
at  D.  From  any  point  G  of  the 
curve  draw  GKG'  parallel  to  TT' 
and  cutting  DD'  in  K ;  then  is 
GK  an  ordinate  to  the  diameter 
DD'. 

It  is  proved  in  Prop.  XIX.,  Cor. 
1,  that  GK  is  equal  to  G'K  ;  hence  the  entire  line  GG'  is  call- 
ed a  double  ordinate. 

11.  The  parts  into  which  a  diameter  is  divided  by  an  or- 
dinate, are  called  abscissas. 

Thus,  DK  and  D'K  are  the  abscissas  of  the  diameter  DD' 
corresponding  to  the  ordinate  GK. 

12.  Two  diameters  are  conjugate  to  one  another,  when 
each  is  parallel  to  the  ordinates  of  the  other. 

Thus,  draw  the  diameter  EE'  parallel  to  GK,  an  ordinate 
to  the  diameter  DD',  in  which  case  it  will,  of  course,  be  par- 
allel to  the  tangent  TT' ;  then  is  the  diameter  EE'  conju- 
gate to  DD'. 

13.  The  latus  rectum  is  the  double  ordinate  to  the  major 
axis  which  passes  through  one  of  the  foci. 

Thus,  through  the  focus  F' 
draw  LL/  a  double  ordinate  to 
the  major  axis,  it  will  be  the  latus 
rectum  of  the  ellipse. 

14.  A  subtangent  is  that  part 
of  the  axis  produced  which  is  in- 
cluded between  a  tangent  and  the 
ordinate  drawn  from  the  point  of 
contact. 

Thus,  if  TT'  be  a  tangent,  to  the  curve  at  D,  and  DG  an 
ordinate  to  the  major  axis,  then  GT  is  the  correspond  ing 
subtangeiit. 

15.  If  a  tangent,  LT,  to  the 
ellipse  be  drawn  through  one 
extremity  of  the  latus  rectum, 
LL',  meeting  the  axis  produced 
in  T,  a  straight  line,  GT,  drawn 
through  the  point  of  intersec- 
tion perpendicular  to  the  axis, 

is  called  the  directrix  of  the  ellipse. 


190 


CONIC    SECTIONS. 


PROPOSITION   I.       PROBLEM. 

To  describe  an  ellipse. 

Let  F  and  F'  be  any  two  fixed 
points.  Take  a  thread  longer  .than 
the  distance  FF',  and  fasten  one  of 
its  extremities  at  F,  the  other  at  F'. 
Then  let  a  pencil  be  made  to  glide 
along  the  thread  so  as  to  keep  it  al- 
ways stretched ;  the  curve  described 
by  the  point  of  the  pencil  will  be  an 
ellipse.  For,  in  every  position  of  the 
pencil,  the  sum  of  the  distances  DF,  DF'  will  be  the  same, 
viz.,  equal  to  the  entire  length  of  the  string. 


PROPOSITION   II.       THEOREM. 

-v 

The  sum  of  the  two  lines  drawn  from  any  point  of  an  ellipse, 
to  the  foci,  is  equal  to  the  major  axis. 

Let  ADA'  be  an  ellipse,  of 
which  F,  F/  are  the  foci,  AA'  is 
the  major  axis,  and  D  any  point  of 
the  curve ;  then  will  DF+DF'  be 
equal  to  AA'. 

For,  by  Def.  1,  the  sum  of  the 
distances  of  any  point  of  the  curve 
from  the  foci,  is  equal  to  a  given  line.  Now,  when  the  point 
D  arrives  at  A,  FA+F'A  or  2AF+FF'  is  equal  to  the  given 
line.  And  when  D  is  at  A',  FA'+F'A'  or  2A'F>+FF'  is 
equal  to  the  same  line.  Hence 

2AF+FF'  =  2A'F'+FF' ; 
consequently,          AF  is  equal  to  A'F'. 
Hence  DF+DF',  which  is  equal  to  AF+AF',  must  be  equal 
to  AA'.     Therefore,  the  sum  of  the  two  lines,  &c. 

Cor.  The  major  axis  fo  bisected  in  the  center.  For,  by  Def. 
3,  CF  is  equal  to  CF' ;  and  we  have  just  proved  that  AF  is 
equal  to  A'F' ;  therefore  AC  is  equal  to  A'C. 


ELLIPSE. 


191 


PROPOSITION    III.       THEOREM 

Every  diameter  is  bisected  in  the  center. 

Let  D  be  any  point  of  an  ellipse ; 
join  DF,  DF',  and  FF'.  Complete  the 
parallelogram  DFD'F',  and  join  DD'. 

Now,  because  the  opposite  sides  of 
a  parallelogram  a-re  equal,  the  sum  of 
DF  and  DF'  i«  equal  to  the  sum  of 
D'F  and  D'F'  *  hence  D'  is  a  point  in 
the  ellipse.  But  the  diagonals  of  a  parallelogram  bisect  each 
other ;  therefore  FF'  is  bisected  in  C  ;  that  is,  C  is  the  center 
of  the  ellipse,  and  DD'  is  a  diameter  bisected  in  C.  There- 
fore, every  diameter,  &c. 


PROPOSITION   IV.       THEOREM. 

The  distance  from  either  focus  to  the  extremity  of  the  minor 
axis,  is  equal  to  half  the  major  axis. 

Let  F  and  F'  be  the  foci  of  an 
ellipse,  AA'  the  major  axis,  and 
BB'  the  minor  axis ;  draw  the 
straight  lines  BF,  BF' ;  then  BF, 
BF'  are  each  equal  to  AC. 

In  the  two  right-angled  trian- 
gles BCF,  BCF',  CF  is  equal  to 
CF',  and  BC  is  common  to  both 
triangles  ;  hence  BF  is  equal  to  BF'.  But  BF+BF'  is  equal 
to  2AC  (Prop.  II.);  consequently,  BF  and  BF'  are  each 
equal  to  AC.  Therefore,  the  distance,  &c. 

Cor.  1.  Half  the  minor  axis  is  a  mean  proportional  between 
the  distances  from  either  focus  to  the  principal  vertices. 

For  BC2  is  equal  to  BF'-FCa  (Prop.  XL,  B.  IV.),  which 
is  equal  to  AC2— FC«  (Prop.  IV.).     Hence  (Prop.  X.,  B.  IV.), 
BC'  =  (AC+FC)  x(AC-FC) 
=  AF'xAF;  and,  therefore, 
AF  :  BC  :  :  BC  :  FA'. 

Cor.  2.   The  square  of  the  eccentricity  is  equal  to  the  differ- 
ence of  the  squares  of  the  semi-axes. 

For  FC3  is  equal  to  BF2— BC2,  which  is  equal  to  ACa— 
BC2. 


192  CONIC    SECTIONS. 


PROPOSITION   V.       THEOREM. 

A  tangent,  to  the  ellipse  makes  equal  angles  with  straight 
''ines  drawn  from  the  point  of  contact  to  the  foci. 

Let  F,  F'  be  the  foci  of  an  ellipse, 
and  D  any  point  of  the  curve ;  if 
through  the  point  D  the  line  TT' 
be  drawn,  making  the  angle  TDF 
equal  to  T'DF',  then  will  TT'  be 
a  tangent  to  the  ellipse  at  D. 

For  if  TT'  be  not  a  tangent,  it 
must  meet  the  curve  in  some  other 
point  than  D.  Suppose  it  to  meet  the  curve  in  the  point  E. 
Produce  F'D  to  G,  making  DG  equal  to  DF ;  and  join  EF, 
EF',  EG,  and  FG. 

Now,  in  the  two  triangles  DFH,  DGH,  because  DF  is  equal 
to  DG,  DH  is  common  to  both  triangles,  and  the  angle  FDH 
is,  by  supposition,  equal  to  F'DT',  which  is  equal  to  the  ver- 
tical angle  GDH ;  therefore  HF  is  equal  to  HG,  and  the  an- 
gle DHF  is  equal  to  the  angle  DHG.  Hence  the  line  TT' 
is  perpendicular  to  FG  at  its  middle  point ;  and,  therefore, 
EF  is  equal  to  EG. 

Also,  F'G  is  equal  to  F'D+DF,  or  F'E+EF,  from  the  na- 
ture of  the  ellipse.  But  F'E+EG  is  greater  than  F'G  (Prop. 
VIII.,  B.  I.)  ;  it  is,  therefore,  greater  than  F'E+EF.  Con- 
sequently EG  is  greater  than  EF;  which  is  impossible,  for 
we  have  just  proved  EG  equal  to  EF.  Therefore  E  is  not 
a  point  of  the  curve,  and  TT'  can  not  meet  the  curve  in  any 
other  point  than  D ;  hence  it  is  a  tangent  to  the  curve  at  the 
point  D.  Therefore,  a  tangent  to  the  ellipse,  &c. 

Cor.  1.  The  tangents  at  the  vertices  of  the  axes,  are  per- 
pendicular to  the  axes ;  and  hence  an  ordinate  to  either  axis 
is  perpendicular  to  that  axis. 

Cor.  2.  If  TT'  represent  a  plane  mirror,  a  ray  of  light 
proceeding  from  F  in  the  direction  FD, would  be  reflected  in 
the  direction  DF',  making  the  angle  of  reflection  equal  to  the 
angle  of  incidence.  And,  since  the  ellipse  may  be  regarded 
as  coinciding  with  a  tangent  at  the  point  of  contact,  if  rays 
of  light  proceed  from  one  focus  of  a  concave  ellipsoidal  mir- 
ror, they  will  all  be  reflected  to  the  other  focus.  For  this 
reason,  the  points  F,  F'  are  called  the/oci,  or  burning  points. 


ELLIPSE.  193 


PROPOSITION    VI.       THEOREM. 

Tangents  to  the  ellipse  at  the  vertices  of  a  diameter,  are  pai< 
allel  to  each  other. 

Let  DD  be  any  diame- 
ter of  an  ellipse,  and  TT' 
W  tangents  to  the  curve 
at  the  points  D,  D' ;  then 
will  they  be  parallel  to  each  _,, 
other. 

Join  DF,  DF',  D'F,  D'F' ; 
then,  by  the  preceding  Prop- 
osition, the  angle  FDT  is 
equal  to  F'DT',  and  the  an- 
gle  FD'V  is  equal  to  F'D'V'.  But,  by  Prop.  III.,  DFD'F'  is 
a  parallelogram ;  and  since  the  opposite  angles  of  a  parallelo- 
gram are  equal,  the  angle  FDF'  is  equal  to  FD'F' ;  therefore 
the  angle  FDT  is  equal  to  F'D'V'  (Prop.  II.,  B.  I.).  Also, 
since  FD  is  parallel  to  F'D',  the  angle  FDD'  is  equal  tc 
F'D'D ;  hence  the  whole  angle  D'DT  is  equal  to  DD'V ; 
and,  consequently,  TT'  is  parallel  to  VV.  Therefore,  tan- 
gents, &c. 

Cor.  If  tangents  are  drawn  through  the  vertices  of  any 
two  diameters,  they  will  form  a  parallelogram  circumscribing 
the  ellipse. 


PROPOSITION  VII.       THEOREM. 


If  from  the  vertex  of  any  diameter,  straight  lines  are  drawn 
through  the  foci,  meeting  the  conjugate  diameter,  the  part  in 
tercepted  by  "the  conjugate  is  equal  to  half  the  major  axis. 

Let  EE'  be  a  diameter  conju- 
gate to  DD',  and  let  the  lines  DF, 
DF'  be  drawn,  and  produced,  if 
necessary,  so  as  to  meet  EE'  in 
II  and  K ;  then  will  DH  or  DK 
be  equal  to  AC. 

Draw  FG  parallel  to  EE'  or 
TT'.     Then  the  angle  DGF  is 
equal    to     the    alternate     angle 
F'DT',  and  the  angle  DFG  is  equal  to  FDT.     But  the  angles 
FDT,  F'DT'  are  equal  to  each  olhe,  (Prop.  V.) ;  hence  the 


194  CLNIC  SECTIONS. 

angles  DGF,  DFG  are  equal  to  each  other,  and  DG  is  equa. 
to  DF.  Also,  because  CH  is  parallel  to  FG,  and  CF  is  equa 
to  CF' ;  therefore  HG  must  be  equal  to  HF'. 

Hence  FD+F'D  is  equal  to  2DG+2GH  or  2DH.  But 
FD+F'D  is  equal  to  2AC.  Therefore  2AC  is  equal  to  2DH, 
or  AC  is  equal  to  DH. 

Also,  the  angle  DHK  is  equal  to  DKH  ;  and  her»ce  DK  is 
equal  to  DII  or  AC.  Therefore,  if  from  the  vertex.  &c. 


PROPOSITION   V11I.       THEOREM. 

Perpendiculars  drawn  from  the  foci  upon  a  tangent  to  the 
ellipse,  meet  the  tangent  in  the  circumference  of  a  circle,  whose 
diameter  is  the  major  axis. 

Let  TT'  be  a  tangent  to  the 
ellipse  at  D,  and  from  F'  draw 
F'E  perpendicular  to  T'T  ;  the 
point  E  will  be  in  the  circum- 
ference of  a  circle  describe^! 
upon  AA'  as  a  diameter. 

Join  CE,  FD,  F'D,  and  pro- 
duce F'E  to  meet  FD  produced 
inG. 

Then,  in  the  two  triangles 
DEF',  DEG,  because  DE  is  com- 
mon to  both  triangles,  the  angles 
at  E  are  equal,  being  right  angles ;  also,  the  angle  EDF'  i? 
equal  to  FDT  (Prop.  V.),  which  is  equal  to  the  vertical  an- 
gle EDG ;  therefore  DF'  is  equal  to  DG,  and  EF'  is  equal 
to  EG. 

Also,  because  F'E  is  equal  to  EG,  and  F'C  is  equal  to  CF, 
CE  must  be  parallel  to  FG,  and,  consequently,  equal  to  half 
ofFG. 

But,  since  DG  has  been  proved  equal  to  DF',  FG  is  equal 
to  FD+DF',  which  is  equal  to  AA'.  Hence  CE  is  equal  to 
half  of  AA'  or  AC  ;  and  a  circle  describe^  with  C  as  a  cen- 
ter, and  radius  CA,  will  pass  through  the  point  E.  The  same 
may  be  proved  of  a  perpendicular  let  fall  upon  TT'  from  the 
focus  F.  Therefore,  perpendiculars,  &c. 

Cor.  CE  is  parallel  to  DF,  and  if  CH  be  joiner ,  CH  will 
be  parallel  o  DF7. 


ELLIFSE. 


195 


PROPOSITION  IX.       THEOREM. 

The  product  of  the  perpendiculars  from  the  foci  ufon  a  tan 
gent,  is  equal  to  the  square  of  half  the  minor  axis. 

Let  TT'  be  a  tangent  to  the 
ellipse  at  any  point  E,  and  let 
the  perpendiculars  FD,  F'G  be 
drawn  from  the  foci ;  then  will 
the  product  of  FD  by  F'G,  be 
equal  to  the  square  of  BC. 

On  AA',  as  a  diameter,  de- 
scribe a  circle ;  it  will  pass 
hrough  the  points  D  and  G 
(Prop.  VIII.).  Join  CD,  and 
produce  it  to  meet  GF'  in  D'. 
Then,  because  FD  and  F'G  are  perpendicular  to  the  same 
straight  line  TT',  they  are  parallel  to  each  other,  and  the  al- 
ternate angles  CFD,  CF'D'  are  equal.  Also,  the  vertical 
angles  DCF,  D'CF'  aie  equal,  and  CF  is  equal  to  CF'. 
Therefore  (Prop.  VII.,  B.  I.)  DF  is  equal  to  D'F',  and  CD  is 
equal  to  CD' ;  that  is,  the  point  D'  is  in  the  circumference  of 
the  circle  ADGA'. 

Hence  DFxGF'  is  equal  to  D'F'xGF',  which  is  equal  to 
A'F'xF'A  (Prop.  XXVIL,  B.  IV.),  which  is  equal  to  BCa 
(Prop.  IV.,  Cor.  1).  Therefore,  the  product,  &c. 

Cor.  The  triangles  FDE,  F'GE  are  similar  ;  hence 

FD  :  F'G  :  :  FE  :  F'E ; 

that  is,  perpendiculars  let  fall  from  the  foci  upon  a  tangent, 
are  to  each  other  as  the  distances  of  the  point  of  contact,  from 
the  foci 


PROPOSITION   X.       THEOREM. 


If  a  tangent  and  ordinate  be  drawn  from  the  same  point  of 
an  ellipse,  meeting  either  axis  produced,  half  of  that  axis  will 
be  a  mean  proportional  between  the  distances  of  the  two  inter 
sections  from  the  center. 

Let  TT'  be  a  tangent  to  the  ellipse,  and  DG  an  ordinate 
to  the  major  axis  from  the  point  of  contact ;  then  we  shall 
have  CT  :  CA  :  :  CA  :  CG. 

Join  DF,  DF' ;  then,  since  the  exterior  ang.e  of  the  trian 
jrle  FDF'  is  bisected  by  DT  (Prop.  V.),  we  have 


196 


FT  :  FT  :  :  FD  :  FD  (Prop.  XVIL,  Sch.,  B.  IV). 
Hence,  by  Prop.  VII,  Cor.,  B.  II., 

F'T+FT  :  FT— FT  :  :  F'D+FD  :  F'D-FD, 
or  2CT  :    FF  :  :  2CA  :  F'D-FD  ;      9 

that  is,        2CT  :  2CA  :  :    FF  :  FD-FD.  (1) 

Again,  because  DG  is  drawn  from  the  vertex  of  the  trian* 
gle  FDF  perpendicular  to  the  base  FF',  we  have  (Prop. 
XXXI.,  Cor.,  B.  IV.), 

FF  :  FD— FD  :  :  FD+FD  :  FG— FG, 
or  FF  :  F'D-FD  :  :  2CA  :  2CG.          0         (2) 

Comparing  proportions  (1)  and  (2),  we  have 
2CT  :  2CA  :  :  2CA  :  2CG, 


or 


CT:    CA 

It  may  also  be  proved  that 

CT'  :  CB 
Therefore,  if  a  tangent,  &c. 


CA:    CG. 
CB  :  OG'. 


; 


PROPOSITION    XI.       THEOREM. 


The  subtangent  of  an  ellipse,  is  equal  to  the  corresponding 
subtangent  of  the  circle  described  upon  its  major  axis. 


Let  AEA'  be  a  circle  de- 
scribed on  AA',  the  major 
axis  of  an  ellipse ;  and  from 
any  point  E  in  the  circle, 
draw  the  ordinate  EG  cut- 
the  ellipse  in  D.  Draw 
touching  the  ellipse  at 
join  ET;  then  will  ET 


E 


ii  a  tangent  to  the  circle  at  E. 
Toin  CE.     Then,  by  the  last  Proposition, 

CT  :  CA  :  :  CA  :  CG ; 
or,  because  CA  is  equal  to  CE, 

CT  :  CE  :  :  CE  :  CG 

Hence  the  triangles  GET,  CGE,  having  the  angle  at  C  com 
Don,  and  the  sides  about  this  angle  proportional,  are  similar 
Therefore  the  angle  CET,  being  equal  to  the  angle  CGE,  is 


ELLIPSE. 


191 


a  right  angle ;  that  is,  the  line  ET  is  perpendicular  to  the 
radius  CE,  and  is,  consequently,  a  tangent  to  the  circle  (Prop. 
IX.,  B.  III.).  Hence  GT  is  the  subtangent  corresponding  to 
each  of  the  tangents  DT  and  ET.  Therefore,  the  subtan- 
gent, &c. 

Cor.  A  similar  property  may  bo  proved  of  a  tangent  to  the 
ellipse  meeting  the  minor  axis.  3> 


PROPOSITION   XII.       THEOREM. 

The  square  of  either  axis,  is  to  the  square  of  the  other,  as  tha 
rectangle  of  the  abscissas  of  the  former,  is  to  the  square  of  their 
ordinate. 


Let  DE  be  an  ordinate  to  the 

major  axis  from  the  point  D ; 

then  we  shall  have 

CA2:CB2::  AExEA':DE2. 

Draw  TT'  a  tangent  to  the 

ellipse  at  D,  then,  by  Prop.  X., 

CT  :  CA  :  :  CA  :  CE.. 
Hence  (Prop.  XII.,  B.  II.), 

CA8  :  CE2  :  :  CT  :  CE ; 
and,  by  division  (Prop.  VII.,  B. 
II.),  CA2  :  CA2-CE2 

Again,  by  Prop.  X., 

CT'  :  CB  :  :  CB 
Hence  (Prop.  XII.,  B.  II.), 

CB2  :  DE2  :  : 
But,  by  similar  triangles, 

CT'  :  DE 
therefore  CB2  :  DE2 


:  :  CT  :  ET. 

:  CE'  or  DE. 
CT/  :  DE. 


(2) 


CT : ET ; 
CT  :  ET. 

Comparing  proportions  (1)  and  (2),  we  have 

CA2  :  CA2-CE2  :  :  CB2  :  DE2. 
But  CA2— CE2  is  equal  to  AExEA'  (Prop.  X.,  B.  IV.); 
hence  CA2  :  CB2  :  :  AE  xEA'  :  DEa. 

In  the  same  manner  it  may  be  proved  that 

CB2  :  CA2  :  :  BE'xE'B'  :  DE'2. 
Therefore,  the  square,  &c. 

Cor.  1.  CA2  :  CB2  :  :  CA2-CE2  :  DE;. 

Cor.  2.  The  squares  of  the  ordinates  to  either  axis,  are  to 
each  other  as  the  rectangles  of  their  abscissas. 

Cor.  3.  If  a  circle  be  described  on  either  axis,  then  any  or- 
dinate in  the  circle,  is  to  the  corresponding  ordinate  in  th& 
ellipse,  as  the  axis  of  that  ordinate,  is  to  the  other  aris. 


198 


CONIC    SECTIONS. 


For,  by  the  Proposition, 
CAa  :  CB2  :  :  AE  xEA'  :  DE2. 
But  AE  X  EA'   is  equal   to  GE2 
(Prop.  XXIL,  Cor.,  B.  IV.). 
Therefore  CA8 :  CB2  :  :  GE2  :  DE2, 
or        CA :  CB  :  :  GE  :  DE. 
In-  the   same   manner  it  may  be 
proved  that 

CB  :  CA  :  :  G'E'  :  DE'.      <D 


PROPOSITION   XIII.       THEOREM. 


The  latus  rectum  is  a  third  proportional  to  the  major  ana 
minor  axes. 


Let  LL'  be  a  double  ordinate  to 
the  major  axis  passing  through  the 
focus  F  ;  then  we  shall  have 

AA'  :  BB'  :  :  BB'  :  LL'. 
Because  LF  is  an  ordinate  to  the 


major  axis, 
AC2 


BCa 


AFxFA':LF2  (Prop 
:  :  BC2  :  LF2  (Prop.  IV., 
Hence  AC  :  BC   :  :  BC  :  LF, 

or  AA' :  BB'  :  :  BB'  :  LL'. 

Therefore,  the  latus  rectum,  &c. 


B' 
.  XII.). 

Cor.  1). 


PROPOSITION    XIV.       THEOREM. 

If  from  the  vertices  of  two  conjugate  diameters,  ordinates  are 
drawn  to  either  axis,  the  sum  of  their  squares  will  be  equal  to 
the  square  of  half  the  other  axis. 

Let  DD',  EE'  be  any 
two  conjugate  diameters, 
DG  and  EH  ordinates  to 
the  major  axis  drawn 
from  their  vertices ;  in 
which  case,  CG  and  CH 
will  be  equil  to  the  ordi- 
nates to  the  minor  axis 
drawn  from  the  same  points  ;  then  we  shall  have 
CA2  =  CG2+CH2,  and  CB2  =  DG2+EH2. 

Let  DT  be  a  tangent  to  the  ellipse  at  D,  and  ET'  a  tan 
(jent  at  E.     Then,  by  Prop.  X., 


ELLIPSE. 


whence      CG  :  CH 


CGxCT  is  equal  to  CA2,  or  CHxCT'; 


:  CT'  :  CT  ;  or,  by  similar  triangles, 
:  CE   :  DT  ;  that  is, 
:  CH  :  GT. 


Hence 


CH2=  GT  XCG, 


=  (CT— CG)XCG 
=  CG  xCT-CG3 
=  CAa-CGa  (Prop.  X.); 
that  is,  CAa=  CGa+CHa. 

In  t.ie  same  manner  it  may  be  proved  that 

CB2=DG2+EH2. 
Therefore,  if  from  the  vertices,  &c. 

Cor.  1.    CHa  is  equal  to  CA2— CG8;   that   is,  CGxGT; 
hence  (Prop.  XII.,  Cor.  1), 

CAa:CB2::CGxGT  :  DG2. 

Cor.  2.  CGa  is  equal  to  CA2— CH2  or  AHxHA' ;  hence 
CAa  .  CBa  :  :  CG2  :  EH2. 


PROPOSITION  XV.       THEOREM. 

The  sum  of  the  squares  of  any  two  conjugate  diameters,  rs 
equal  to  the  sum  of  the  squares  of  the  axes. 

Let  DD',  EE'  be  any  two  con- 
jugate diameters ;  then  we  shall 
have 

DD'2+EE'2=AA'2+BB'2. 

Draw  DG,  EH  ordinates  to  the 
major  axis.  Then,  by  the  prece- 
ding Proposition, 

CG2+CH2=CA2, 
and  DG2+EH2  =  CB2. 

Hence  CG2+DG2+CH2-fEH9  =  CA'+CB2, 

or  CD2+CE2=CA2+CB2 ; 

that  is,  DD'2+ EE'a=AA'2+BB'a. 

Therefore,  the  sum  of  the  squares,  &c. 


PROPOSITION   XVI.       THEOREM. 

The  parallelogram  formed  by  drawing  tangents  through  tnc 
vertices  of  two  conjugate  diameters,  is  equal  to  the  rectangle  of 
the  axes. 

Let  DED'E'  be  a  parallelogram,  formed  by  drawing  tan 
gents  to  the  e^ipse  through  the  vertices  of  two  conjugate 
diameters  DD',  EEf ;  its  area  is  equal  to  AA'xBB'. 


200 


CONIC    SECTIONS. 


>T 


Let  the  tangent  at  D,  meet  the  major  axis  produced  in  T  i 
join  E'T,  and  draw  the  ordinates  DG,  E'H. 

Then,  by  Prop.  XIV.,  Cor.  2,  we  have 

CA2  :  CBa  :  :  CGa    E'H2, 
or  CA  :  CB  :  :  CG     E'H. 

But  CT  :  CA  :  :  CA     CG  (Prop.  X.) ; 

hence  CT  :  CB  :  :  CA     E'H, 

or  CA  x  CB  is  equal  to  CT  x  E'H, 

which  is  equal  to  twice  the  triangle  CE'T,  or  the  parallelo 
gram  DE' ;  since  the  triangle  and  parallelogram  have  the 
same  base  CE',  and  are  between  the  same  parallels. 

Hence  4CAxCB  or  AA'xBB',  is  equal  to  4DE',  or  the 
parallelogram  DED'E'.     Therefore,  the  parallelogram,  &c. 


PROPOSITION    XVII.       TIIEGHEM. 

If  from  the  vertex  of  any  diameter,  straight  lines  are  drawn 
to  the  foci,  their  product  is  equal  to  the  square  of  half  the  con- 
jugate diameter. 

Let  DD',  EE'  be  two  conjugate 
diameters,  and  from  D  let  lines 
be  drawn  to  the  foci ;  then  will 
FDxF'D  be  equal  to  ECa. 

Draw  a  tangent  to  the  ellipse 
at  D,  and  upon  it  let  fall  the  per- 
pendiculars  FG,  F'H  ;  draw,  also, 
DK  perpendicu.ar  to  EE'. 

Then,  because  the  triangles 
DFG,  DLK,  DF'H  are  similar,  we  have 

FD:FG  :  :DL:DK. 
Also,  F'D  :  PH  :  :  DL  :  DK. 

Whence  (Prop.  XL,  B.  II.), 

FDxF'D  :  FGxF'H  :  :  DL2  :  DK2.  (1\ 

But,  by  Prop.  XVI  , ACxBC=ECxDK; 
whence  AC  or  DL  :  DK  :  :  EC  :  BC, 

and  DL2  :  DK2  :  :  ECa :  BC2.  (2) 


ELLIPSE.  201 

Comparing  proportions  (1)  and  (2)  we  have 

FDxF'D  :  FGxF'H  :  :  EC2  :  BCa. 

But  FGxF'H  is  equal  to  BC2  (Prop.  IX.) ;  hence  FDxF'D 
is  equal  to  EC2.     Therefore,  if  from  the  vertex,  &e. 

PROPOSITION    XVIII.       THEOREM. 

If  a  tangent  and  ordinate  be  drawn  from  the  same  point  of 
an  ellipse  to  any  diameter -,  half  of  that  diameter  will  be  a  mean 
proportional  between  the  distances  of  the  two  intersections  from 
the  center. 

Let  a  tangent  EG  and  an  ordinate  EH  be  drawn  from  the 
same  point  E  of  an  ellipse,  meeting  the  diameter  CD  pro- 
;  then  we  shall  have 

CG  :  CD  :  :  CD  :  CH. 


o 

Produce  EG  and  EH  to  meet  the  major  axis  in  K  and  L ; 
•draw  DT  a  tangent  to  the  curve  at  the  point  D,  and  draw 
DM  parallel  to  GK.     Also,  draw  the  ordinates  EN,  DO. 
By  Prop.  XIV.,  Cor.  1,  CA2  :  CB2  :  :  COxOT  :  DO2, 

:CNxNK:EN2. 
Hence 
COxOT  :  CNxNK  :  :  DO2  :  EN2 

:  :  OT2  :  NL2,  by  similar  triangles.     (ly 
Also,  by  similar  triangles,  OT  :  NL  :  :  DO  :  EN 

:  :  OM  :  NK.  (2) 

Multiplying  together  proportions  (1)  and  (2)  (Prop.  XL, 
B.  II.),  and  omitting  the  factor  OT2  in  the  antecedents,  and 
NKxNL  in  the  consequents,  we  have 

CO:CN::OM:NL; 

and,  bj  composition,  CO  :  CN  :  :  CM  :  CL.  (3) 

Also,  by  Prop.  X.,  CKxCN=CA2=CTxCO; 
hence  CO  :  CN  :  :  CK  :  CT.  (4) 

Comparing  proportions  (3)  and  (4),  we  have 

CK  :  CM  :  :  CT  :  CL. 
But  CK  :  CM  :  :  CG  :  CD, 

and  CT  :  CL  :  :  CD  :  CH  ; 

nence  CG  :  CD  :    CD  :  CH. 

Therefore,  if  a  tangent,  &c. 


202  CONIC    SECTIONS. 


PROPOSITION   XIX.       THEOREM. 

The  squan  oj  any  diameter,  is  to  the  square  of  its  conjugate^ 
as  the  rectangle  of  its  abscissas,  is  to  the  square  of  their  or* 
dinate. 

Let  DD',  EE'  be  two  conjugate    T/ 
diameters,  and  GH  an  ordinate  to          "  ~ 
DD';  then 

DD'3  :  EE'2  :  :  DHxHD'  :  GH2. 

Draw   TT'    a   tangent    to    the 

curve  at  the  point  G,  and  draw 

GK  an  ordinate  to  EE'.     Then, 

by  Prop.  XVIII., 

CT  :  CD  :  :  CD 
and  CD2  :  CHa :  :  CT  :  CH  (Prop.  XII.,  B.  II.) ; 

whence,  by  division, 

CD'  :  CD2— CH2  :  :  CT  :  HT.  (1) 

Also,  by  Prop.  XVIII., 

CT'  :  CE  :  :  CE  :  CK, 
and  CE2  :  CK2 :  :  CT' :  CK  or  GH, 

:  :  CT  :  HT.  (2) 

Comparing  proportions  (1)  and  (2),  we  have 

CD2  :  CE2  :  :  CD2-CH2  :  CK2  or  GH2, 
or  DD'2 :  EE'2 :  :  DH  x  HD'  :  GH2. 

Therefore,  the  square,  &c. 

Cor.  1.  In  the  same  manner  it  may  be  proved  that 
DD'2  :  EE'2  :  :  DH  x  HD'  :  G'H* ;  hence  GH  is  equal  to  G'H, 
or  every  diameter  bisects  its  double  ordinates. 

Cor.  2.  The  squares  of  the  ordinates  to  any  diameter,  are 
to  each  other  as  the  rectangles  of  their  abscissas. 


PROPOSITION   XX.       THEOREM. 


^ ") 


If  a  cone  be  cut  by  a  plane,  making  an  angle  with  the  base 
less  than  that  made  by  the  side  of  the  cone,  the  section  is  an 
ellipse. 

Let  ABC  be  a  cone  cut  by  a  plane  DEGH,  making  an  an- 
gle with  the  base,  less  than  that  made  by  the  side  of  the  cone ; 
the  section  DeEGH/i  is  an  ellipse. 

Let  ABC  be  a  section  through  the  axis  of  the  cone,  and 
perpendicular  to  the  plane  DEGH.  Let  EMHO.  emho  be 
circular  sections  parallel  to  the  base  ;  then  EH,  the  intersec- 


ELUPSE. 


tion  of  the  planes  DEGH,  EMHO,  will 
be  perpendicular  to  the  plane  ABC, 
and,  consequently,  to  each  of  the  lines 
DG,  MO.  So,  also,  th  will  be  perpen- 
dicular to  DG  and  mo. 

Now,  because  the  triangles  DNO, 
Dno  are  similar,  as  also  the  triangles 
GMN,  Gmn,  we  have  the  proportions, 

NO  i  no  :  :  DN  :  DTI, 
and  MN  :  mn  :  :  NG  :  nG. 
Hence,  by  Prop.  XL,  B.  II., 

MNxNO  :  mnxno  :  :  DNxNG  :  DnxnG. 
But  since  MO  is  a  diametei  of  the  circle  EMHO,  and  EN  is 
perpendicular  to  MO,  we  have  (Prop.  XXII.,  Cor.,  B.  IV.). 

MNxNO=EN2. 
For  the  same  reason,  mnxno  =  eri*. 

Substituting  these  values  of  MNxNO  and  mnxno,  in  the 
preceding  proportion,  we  have 

EN3  -.en1::  DNxNG  :  DrcxrcG; 

that  is,  the  squares  of  the  ordinates  to  the  diameter  DG,  are 
to  each  other  as  the  products  of  the  corresponding  abscissas. 
Therefore  the  curve  is  an  ellipse  (Prop.  XII.,  Cor.  2)  whose 
major  axis  is  DG.  Hence  the  ellipse  is  called  a  conic  section. 
as  mentioned  on  page  177. 


PROPOSITION   XXI.       THEOREM. 

The  area  of  an  ellipse  is  a  mean  proportional  between 
two  circles  described  on  its  axes. 

Let  AA'  be  the  major  axis  of  an 
ellipse  ABA'B'.  On  AA'  as  a  di- 
ameter, describe  a  circle ;  inscribe 
in  the  circle  any  regular  polygon 
AEDA',  and  from  the  vertices  E, 
D,  &c.,  of  the  polygon,  draw  per- 
pendiculars to  AA'.  Join  the 
points  B,.  G,  &c.,  in  which  these 
perpendiculars  intersect  the  ellipse, 
and  there  will  be  inscribed  in  the 
ellipse  a  polygon  of  an  equal  num- 
ber of  sides. 

Now  the  area  of  the  trapezoid  CEDH,  is  equal  to  (CE-f 

DH)  x-— ;  and  the  area  of  the  trapezoid  CBGH,  is  equal  to 


204 


CONIC   SECTIONS. 


/-l  IT 

vCB+GH)x— .   These trapezoids 

are  to  each  other,  as  CE+DH  to 
CB+GH,  or  as  AC  to  BC  (Prop. 
XII.,  Cor.  3). 

In  the  same  manner  it  may  be 
proved  that  each  of  the  trapezoids 
composing  the  polygon  inscribed  in 
the  circle,  is  to  the  corresponding 
trapezoid  of  the  polygon  inscribed 
in  the  ellipse,  as  AC  to  BC.  Hence, 
the  entire  polygon  inscribed  in  the  circle,  is  to  the  polygon  in 
scribed  in  the  ellipse,  as  AC  to  BC. 

Since  this  proportion  is  true,  whatever  be  the  number  oi 
sides  of  the  polygons,  it  will  be  true  when  the  number  is  in 
definitely  increased  ;  in  which  case  one  of  the  polygons  coin 
cides  with  the  circle,  and  the  other  with  the  ellipse.  Hence 
we  have 

Area  of  circle  :  area  of  ellipse  :  :  AC  :  BC. 

But  the  area  of  the  circle  is  represented  by  TrAC2 ;  hence 
the  area  of  the  ellipse  is  equal  to  TT AC  x  BC,  which  is  a  mean 
proportional  between  the  two  circles  described  on  the  axes. 


i: 


PROPOSITION    XXII.       THEOREM. 

The  distance  of  any  point  in  an  ellipse  from  the  directrix  is  to 
its  distance  from  the  focus  nearest  the  directrix,  in  the  constant 
ratio  of  half  the  major  axis  to  the  eccentricity. 

Let  D  be  any  point  in  the  el- 
lipse ;  let  DGr  be  drawn  perpen- 
dicular to  the  directrix  GrT ;  DE 

erpendicular  to  the  axis ;  and  A' 

et  DF,  DE'  be  drawn  to  the 
two  foci.     Take  H,  a  point  in 
the  axis,  so  that  AH=DF,  and, 
consequently,  HA'^DF' ;  then  CH  is  half  the  difference  be- 
tween A'H  and  AH,  or  DF'  and  DF ;  and  CE  is  half  the 
difference  between  F'E  and  FE. 
By  Prop.  XXXI.,  B.  IV., 

DF'+DF :  FF  ::  F'E-FE :  DF'~DF. 
Dividing  each  term  by  two,     CA :  CF  ::  CE :  CH. 
By  Prop.  X.,  Ellipse,     CA*=CF.CT ;  or  CA :  CF ::  CT :  CA. 
Therefore  CT :  C A : :  CE :  CH. 

Hence,  Prop.  VIL,  B.  II,         CT-CE  :  CA-CH 


or 
that  is, 


CT 

ET:AH::CT:CA::CA 
DG:DF::CA:CF. 


CA, 
CF; 


HYPERBOLA. 

Definitions. 

1.  AN  hyperbola  is  a  plane  curve,  in  which  the  difference 
of  the  distances  of  each  point  from  two  fixed  points,  is  equal 
to  a  given  line. 

2.  The  two  fixed  points  are  called  the  foci. 
Thus,  if  F  and  F'  are  two  fixed 

points,  and  if  the  point  D  moves 
about  F  in  such  a  manner  that  the 
difference  of  its  distances  from  F  and 
F'  is  always  the  same,  the  point  D 
will  describe  an  hyperbola,  of  which 
F  and  F'  are  the  foci. 

If  the  point  D'  moves  about  F'  in 
such  a  manner  that  D'F— D'F'  is 
always  equal  to  DF' — DF,  the  point  D' will  describe  a  sec- 
ond hyperbola  similar  to  the  first.  The  two  curves  are  call- 
ed opposite  hyperbolas. 

3.  The  center  is  the  middle  point  of  the  straight  line  join- 
ing the  foci. 

4.  The  eccentricity  is  the  distance  from  the  center  to  either 
focus. 

Thus,  let  F  and  F'  be  the  foci  of 
two  opposite  hyperbolas.  Draw  the 
line  FF',  and  bisect  it  in  C.  The 
point  C  is  the  center  of  the  hyperbola, 
and  CF  or  CF'  is  the  eccentricity. 

5.  A   diameter   is    a   straight    line 
drawn  through  the  center,  and  termi- 
nated by  two  opposite  hyperbolas. 

6.  The  extremities  of  a  diameter  are 
called  its  vertices. 

Thus,  through  C  draw  any  straight  line  DD'  terminated 
by  the  opposite  curves  ;  DD'  is  a  diameter  of  the  hyperbola ; 
D  and  D'  are  its  vertices. 

7.  The  major  axis  is  the  diameter  which,  when  produced, 
passes  through  the  foci ;  and  its  extremities  are  called  the 
principal  vertices. 

8.  The  minor  axis  is  a  line  drawn  through  the  center  per- 


206 


COMO    SECTIONS. 


pendicular  to  the  major  axis,  and  terminated  by  the  circum- 
ference described  from  one  of  the  principal  vertices  as  a  cen- 
ter, and  a  radius  equal  to  the  eccentricity. 

Thus,  through  C  draw  BB'  perpendicular  to  AA',  and  with 
A  as  a  center,  and  with  CF  as  a  radius,  describe  a  circum- 
ference cutting  this  perpendicular  in  B  and  B' ;  then  AA'  is 
the  major  axis,  and  BB'  the  minor  axis. 

If  on  BB'  as  a  major  axis,  opposite  hyperbolas  are  de- 
scribed, having  AA'  as  their  minor  axis,  these  hyperbolas  are 
said  to  be  conjugate  to  the  former. 

9.  A  tangent  is  a  straight  line  which  meets  the  curve,  but, 
being  produced,  does  not  cut  it. 

10.  An  ordinate  to  a  diameter,  is  a  straight  line  drawn 
from  any  point  of  the  curve  to  meet  the  diameter  produced, 
parallel  to  the  tangent  at  one  of  its  vertices. 

Thus,  let  DD'  be  any  diame- 
ter, and  TT'  a  tangent  to  the 
hyperbola  at  D.  From  any 
point  G  of  the  curve  draw 
GKG'  parallel  to  TT'  and  cut- 
ting DI)'  produced  in  K ;  then 
is  GK  an  ordinate  to  the  di- 
ameter DD'. 

It  is  proved,  in  Prop.  XIX., 
Cor.  1,  that  GK  is  equal  to 
G'K ;  hence  the  entire  line  GG'  is  called  a  double  ordinate. 

11.  The  parts  of  the  diameter  produced,  intercepted  be 
tween  its  vertices  and  an  ordinate,  are  called  its  abscissas. 

Thus,  DK  and  D'K  are  the  abscissas  of  the  diameter  DD' 
corresponding  to  the  ordinate  GK. 

12.  Two  diameters  are  conjugate  to  one  another,  when 
each  is  parallel  to  the  ordinates  of  the  other. 

Thus,  draw  the  diameter  EE'  parallel  to  GK  an  ordinate 
to  the  diameter  DD',  in  which  case  it  will,  of  course,  be  par- 
allel to  the  tangent  TT' ;  then  is 
the  diameter  EE'  conjugate  to  DD'. 

13.  The  latus  rectum  is  the  double 
ordinate  to  the   major  axis   which 
passes  through  one  of  the  foci. 

Thus,  through  the  focus  F'  draw 
LL/  a  double  ordinate  to  the  major 
axis,  it  will  be  the  latus  rectum  of 
the  hyperbola 


HYPERBOLA  207 

15.  A  subtangent  is  that  part  of  the  axis  produced  which 
is  included  between  a  tangent,  and  the  ordinate  drawn  from  the 
point  of  contact. 

Thus,  if  TT'  be  a  tangent  to  the  curve  at  D,  and  DG  an 
ordinate  to  the  major  axis,  then  GT  is  the  corresponding 
sub  tangent. 


PROPOSITION   I.       PROBLEM. 

To  describe  an  hyperbola. 

Let  F  and  F'  be  any  two  fixed 
points.  Take  a  ruler  longer  than 
the  distance  FF',  and  fasten  one 
of  its  extremities  at  the  point  F'. 
Take  a  thread  shorter  than  the 
ruler,  and  fasten  one  end  of  it  at 
F,  and  the  other  to  the  end  H  of  the 
ruler.  Then  move  the  ruler  HDF' 
about  the  point  Fr,  while  the  thread  is  kept  constantly  stretched 
by  a  pencil  pressed  against  the  ruler ;  the  curve  described  by 
the  point  of  the  pencil,  will  be  a  portion  of  an  hyperbola. 
For,  in  every  position  of  the  ruler,  the  difference  of  the  lines 
DF,  DF'  will  be  the  same,  viz.,  the  difference  between  the 
length  of  the  ruler  and  the  length  of  the  string. 

If  the  ruler  be  turned,  and  move  on  the  other  side  of  the 
point  F,  the  other  part  of  the  same  hyperbola  may  be  de- 
scribed. Also,  if  one  end  of  the  ruler  be  fixed  in  F,  and  that 
of  the  thread  in  F',  the  opposite  hyperbola  may  be  described. 


PROPOSITION    II.       THEOREM. 

The  difference  of  the  two  lines  drawn  from  any  point  of  an 
hyperbola  to  the  foci,  is  equal  to  the  major  axis. 

Let  F  and  F'  be  the  foci  of  two 
opposite  hyperbolas,  AA'  the  major 
axis,  and  D  any  point  of  the  curve ; 
.hen  will  DF'-DF  be  equal  to  A  A'. 

For,  by  Def.  1,  the  difference  of  the 
distances  of  any  point  of  the  curve 
from  the  foci,  is  equal  to  a  given  line. 
Now  when  the  point  D  arrives  at  A, 
F'A— FA,  or  AA'+F'A'— FA,  is  equal  to  the  given  lino. 
And  when  D  is  at  A',  FA'— PA',  or  AA'-*-AF  —  A'F',  i§ 
equal  to  the  same  line.  Hence 


208 


CONIC   SECTIONS. 


AA'-rAF-A'F'=AA'+F'A'-FA, 
or  2AF=2A'F'; 

that  Ls,  AF  is  equal  to  A'F'. 

Hence  DF'— DF,  which  is  equal  t  AF'— AF,  must  be 
equal  to  A  A'.  Therefore,  the  difference  of  the  two  lines,  &c, 

Cor.  The  major  axis  is  bisected  in  the  center.  For,  by 
Def.  3,  CF  is  equal  to  CF' ;  and  we  have  just  proved  that 
AF  is  equal  to  A'F' ;  therefore  AC  is  equal  to  A'C. 


PROPOSITION   III.       THEOREM. 

Every  diameter  is  bisected  in  the  center. 

Let  D  be  any  point  of  an  hyper- 
bola ;  join  DF,  DF',  and  FF'.  Com- 
plete the  parallelogram  DFD'F',  and 
join  DD'. 

Now,  because  the  opposite  sides  of 
a  parallelogram  are  equal,  the  differ- 
ence between  DF  and  DF'  is  equal 
to  the  difference  between  D'F  and 
D'F' ;  hence  D'  is  a  point  in  the  opposite  hyperbola.     But 
the  diagonals  of  a  parallelogram  bisect  each  other ;  there- 
fore FF'  is  bisected  in  C ;  that  is,  C  is  the  center  of  the  hy 
perbola,  and  DD'  is  a  diameter  bisected  in  C.     Therefore, 
^very  diameter,  &c. 


PROPOSITION    IV.       THEOREM. 

Half  the  minor  axis  is  a  mean  proportional  between  the  dis- 
tances from  either  focus  to  the  principal  vertices. 

Let  F  and  F'  be  the  foci  of  opposite 
hyperbolas,  AA'  the  major  axis,  and  BB' 
the  minor  axis ;  then  will  BC  be  a  mean 
proportional  between  AF  and  A'  F/ 

Join  AB.  Now  BCa  is  equal  to  ABa  — 
AC3,  which  is  equal  to  FC2— AC3  (Def. 
8).  Hence  (Prop.  X.,  B.  IV.), 

BC'=(FC-AC)x(FC+AC) 

=  AFxA'F; 
and  hence  AF  :  BC  :  :  BC  :  A'F. 

Cor.  The  square  of  the  eccentricity  is  equal  to  the  sum  of  th* 
squares  of  the  semi-axes. 

For  FC3  is  equal  to  AB3  (Def.  8),  which  is  equal  to  AC'-f 
BCa. 


HYPERBOLA.  SOD 


PROPOSITION    V.       THEOREM. 

A  tangent  to  the  hyperbola  bisects  the  angle  contained  by 
lines  drawn  from  the  point  of  contact  to  the  foci. 

Let  F,  F'  be  the  foci  of  two 
opposite  hyperbolas,  and  D  any 
point  of  the  curve  ;  if  through  the 
point  D,  the  line  TT'  be  drawn 
bisecting  the  angle  FDF' ;  then 
will  TT'  be  a  tangent  to  the  hy- 
perbola at  D. 

For  if  TT'  be  not  a  tangent,  let 
it  meet  the  curve  in  some  other 
point,  as  E.  Take  DG  equal  to 
DF ;  and  join  EF,  EF',  EG,  and  FG. 

Now,  in  the  two  triangles  DFH,  DGH,  because  DF  is 
equal  to  DG,  DH  is  common  to  both  triangles,  and  the  angle 
FDH  is,  by  supposition,  equal  to  GDH ;  therefore  HF  is 
equal  to  HG,  and  the  angle  DHF  is  equal  to  the  angle  DHG. 
Hence  the  line  TT'  is  perpendicular  to  FG  at  its  middle 
point ;  and,  therefore,  EF  is  equal  to  EG. 

Now  F'G  is  equal  to  F'D— DF,  or  F'E-EF,  from  the 
nature  of  the  hyperbola.  But  F'E— EG  is  less  than  F'G 
(Prop.  VIIL,  B.  I.) ;  it  is,  therefore,  less  than  F'E— EF. 
Consequently,  EG  is  greater  than  EF,  which  is  impossible, 
for  we  have  just  proved  EG  equal  to  EF.  Therefore  E  is 
not  a  point  of  the  curve  ;  and  TT'  can  not  meet  the  curve  in 
any  other  point  than  D ;  hence  it  is  a  tangent  to  the  curve 
at  the  point  D.  Therefore,  a  tangent  to  the  hyperbola,  &c. 

Cor.  1.  The  tangents  at  the  vertices  of  the  axes,  are  per 
pendicular  to  the  axes :  and  hence  an  ordinate  to  either  axis 
is  perpendicular  to  that  axis. 

Cor.  2.  If  TT'  represent  a  plane  mirror,  a  ray  of  light 
proceeding  from  F  in  the  direction  FD,  would  be  reflected  in 
a  line  which,  if  produced,  would  pass  through  F7,  making  the 
angle  of  reflection  equal  to  the  angle  of  incidence.  And, 
since  the  hyperbola  may  be  regarded  as  coinciding  with  a 
tangent  at  the  point  of  contact,  if  rays  of  light  proceed  from 
one  focus  of  a  concave  hyperbolic  mirror,  they  will  be  re- 
flected in  lines  diverging  from  the  other  focus.  For  thii 
reason, the  points  F,  F'  are  called  the  foci. 


210 


CONIC    SECTIONS. 


PROPOSITION    VI.       THEOREM. 

Tangents  to  the  hyperbola  at  the  vertices  of  a  diameter,  are 
parallel  to  each  other. 

Let  DD/  be  any  diameter  of  an 
hyperbola,  and  TT/,  VV/  tangents 
to  the  curve  at  the  points  D,  D' ; 
then  will  they  be  parallel  to  each 
other. 

Join  DF,  DF/,  D'F,  D/F/.  Then, 
by  Prop.  III.,  FDF/D/  is  a  paral- 
lelogram ;  and,  since  the  opposite 
angles  of  a  parallelogram  are  equal, 
the  angle  FDF'  is  equal  to  FD/F/. 
But  the  tangents  TT/,  VV/  bisect  the  angles  at  D  and  D' 
(Prop.  V.) ;  hence  the  angle  F/DT/,  or  its  alternate  angle 
FT'D,  is  equal  to  FD/V.  But  FT'D  is  the  exterior  angle  op- 
posite  to  FD/V ;  hence  TT'  is  parallel  to  VV/.  Therefore 
tangents,  <fcc. 

Cor.  If  tangents  are  drawn  through  the  vertices  of  any 
two  diameters,  they  will  form  a  parallelogram. 


PROPOSITION   VII.       THEOREM. 

If  through  the  vertex  of  any  diameter,  straight  lines  art 
drawn  from  the  foci,  meeting  the  conjugate  diameter,  the  part 
intercepted  by  the  conjugate  is  equal  to  half  of  the  major  axis. 

Let  EE/  be  a  diameter  conjugate  to 
DD/,  and   let   the  lines  DF,  DF/    be  /\ 

drawn,  and  produced,  if  necessary,  so 
as  to  meet  EE/  in  H  and  K  ;  then  will 
DH  or  DK  be  equal  to  AC. 

Draw  F/G  parallel  to  EE/  or  TT/, 
meeting  FD  produced  in  G.  Then  the 
angle  DGF/  is  equal  to  the  exterior 
angle  FDT/ ;  and  the  angle  DF/G  is 
equal  to  the  alternate  angle  F/DT/. 
But  the  angles  FDT/,  F/DT/  are  equal 
to  each  other  (Prop.  V.) ;  hence  the 
angles  DGF',  DF'G  are  equal  to  each  other,  and  DG  is  equa'. 
to  DF/.  Also,  because  CK  is  parallel  to  F/G,  and  CF  is  equa 
to  CF/ ;  therefore  FK  must  be  equal  to  KG. 


HYPERBOLA.  211 


Hence  F'D-FD  is  equal  to  GD  — FD  or  GF— 2DF;  that 
is,  2KF— 2DF  or  SDK.  But  F/D  — FD  is  equal  to  2 AC. 
Therefore  2AC  is  equal  to  2DK,  or  AC  is  equal  to  DK. 

Also,  the  angle  DHK  is  equal  to  DKH  ;  and  hence  DH  is 
equal  to  DK  or  AC.  Therefore,  if  through  the  vertex,  &c. 


PROPOSITION   VIII.       THEOREM. 

Perpendiculars  drawn  from  the  foci  upon  a  tangent  to  t/ie 
hyperbola,  meet  the  tangent  in  the  circumference  of  a  circle 
whose  diameter  is  the  major  axis. 

Let  TT/  be  a  tangent  to  the  hyper- 
bola at  D,  and  from  F  draw  FE  per- 
pendicular to  TT/  ;  the  point  E  will 
be  in  the  circumference  of  a  circle  de- 
scribed upon  AA/  as  a  diameter. 

Join  CE,  FD,  F/D,  and  produce  FE 
to  meet  F/D  in  G. 

Then,  in  the    two   triangles   DEF, 
DEG,  because  DE  is  common  to  both 
triangles,  the  angles  at  E  are  equal,  be- 
ing right  angles ;   also,  the  angle  EDF  is  equal  to  EDG 
(Prop.  V.) ;  therefore  DF  is  equal  to  DG,  and  EF  to  EG. 

Also,  because  FE  is  equal  to  EG,  and  CF  is  equal  to  CF/, 
CE  must  be  parallel  to  F/G,  and,  consequently,  equal  to  half 
ofF'G. 

But,  since  DG  has  been  proved  equal  to  DF,  F/G  is  equal 
to  F/D— FD,  which  is  equal  to  AA/.  Hence  CE  is  equal  to 
half  of  AA/  or  AC ;  and  a  circle  described  with  C  as  a  cen- 
ter, and  radius  CA,  will  pass  through  the  point  E.  The  same 
may  be  proved  of  a  perpendicular  let  fall  upon  TT'  from  the 
focus  F/.  Therefore,  perpendiculars,  &c. 


PROPOSITION   IX.       THEOREM. 

The  product  of  the  perpendiculars  from  the  foci  upon  a  tan- 
gent, is  equal  to  the  square  of  half  the  minor  axis. 

Let  TT/  be  a  tangent  to  the  hyperbola  at  any  point  E, 
and  let  the  perpendiculars  FD,  F/G  be  drawn  from  the  foci ; 
then  will  the  product  of  FD  by  F/G,  be  equal  to  the  square 
ofBC. 

On  AA/  as  a  diameter,  describe  a  circle  ;  it  will  pass 
through  the  points  D  and  G  (Prop.  VIII.) .  Join  CD,  and 


212 


CONIC  SECTIONS. 


produce  it  to  meet  GF'  in  D'.  Then, 
because  FD  and  F'G  are  perpendicu 
lar  to  the  same  straight  line  TT',  they 
are  parallel  to  each  other,  and  the  al- 
ternate angles  CFD,  CF'D'  are  equal. 
Also,  the  vertical  angles  DCF,  D'CF' 
are  equal,  and  CF  is  equal  to  CF'. 
Therefore  (Prop.  VII.,  B.  L),  DF  is 
equal  to  D'F',  and  CD  is  equal  to  C 
that  is,  the  point  D'  is  in  the  circum- 
ference of  the  circle  ADA'G. 

Hence  DFxGF'  is  equal  to  D'F'xGF',  which  is  equal  to 
A'F'XF'A  (Prop.  XXVIII.,  Cor.  2,  B.  IV.),  which  is  equal 
to  BC3  (Prop.  IV.).  Therefore,  the  product,  &c. 

Cor.  The  triangles  FDE,  F'GE  are  similar ;  hence 

FD  :  F'G  :  :  FE  :  F'E ; 

that  is,  perpendiculars  let  fall  from  the  foci  upon  a  tangent,  are 
to  each  other  as  the  distances  of  the  point  of  contact  from  the 
foci. 


an 


PROPOSITION   X.       THEOREM. 

If  a  tangent  and  ordinate  be  drawn  from  the  same  point  oj 
an  hyperbola,  meeting  either  axis  produced,  half  of  that  axis 
will  be  a  mean  proportional  between  the  distances  of  the  two  in- 
tersections from  the  center. 

Let  DTT'  be  a  tangent  to  the 
hyperbola,  and  DG  an  ordinate  to 
the  major  axis  from  the  point  of 
contact ;  then  we  shall  have 
CT  :  CA  :  :  CA  :  CG. 
Join  DF,  DF' ;   then,  since  the 
angle   FDF'    is    bisected   by   DT 
(Prop.  V.),  we  have 

F'T  :  FT  :  :  F'D  ;  FD 
(Prop.  XVIL,  B.  IV.). 
Hence,  by  Prop.  VIL,  Cor.,  B.  II., 

F'T— FT  :  F'T+FT  :  :  F'D— FD 
or  2CT  :   F'F  :  :  2CA :  F'D+FD  ; 

that  is,          2CT  :  2CA  :  :  F'F  :  F'D+FD. 


F'D+FD, 


(1) 

Again,  because  DG  is  drawn  from  the  vertex  of  the  tnan 
gle  FDF'  perpendicular  to  the  base  FF'  produced,  we  have 
(Prop.  XXXI.,  Cor.,  B.  IV.), 

FF  .  F'D+FD  :  .  F'D-FD  :  F'G-f  FG, 
or  F'F  :  F'D+FD  ;  :  2CA  :  2CG.  (2) 


HYPERBOLA.  213 


Comparing  proportions  (1)  and  (2),  we  have 

2CT  :  2CA  :  :  2CA  :  2CG, 
or  CT  :    CA  :  :    CA  :    CG. 

It  may  also  be  proved  that 

CT'  :  CB  :  :  CB  :  CG'. 
Therefore,  if  a  tangent,  &c. 


PROPOSITION   XI.       THEOREM. 

The  subtangent  of  an  hyperbola,  is  equal  to  the  correspond- 
ing subtangent  of  the  circle  described  upon  its  major  axis. 

Let  AEA'  be  a  circle  described  on 
AA'  the  major  axis  of  an  hyperbola ; 
and  from  any  point  E  in  the  circle, 
draw  the  ordinate  ET.  Through  T 
draw  the  line  DT  touching  the  hyper- 
bola in  D,  and  from  the  point  of  con- 
tact draw  the  ordinate  DG.  Join  GE  ; 
then  will  GE  be  a  tangent  to  the  cir- 
cle at  E. 

Join  CE.     Then,  by  the  last  Proposition, 

CT  :  CA  :  :  CA  :  CG ; 
or,  because  CA  is  equal  to  CE, 

CT  :  CE  :  :  CE  :  CG. 

Hence  the  triangles  GET,  CGE  having  the  angle  at  C 
common,  and  the  sides  about  this  angle  proportional,  are  simi- 
lar. Therefore  the  angle  CEG,  being  equal  to  the  angle 
CTE,  is  a  right  angle  ;  that  is,  the  line  GE  is  perpendicular 
to  the  radius  CE,  and.  is,  consequently,  a  tangent  to  the  cir- 
•cle  (Prop.  IX.,  B.  III.).  Hence  GT  is  the  subtangent  cor- 
responding to  each  of  the  tangents  DT  and  EG.  Therefore, 
the  subtangent,  &c. 


PROPOSITION  XII.       THEOREM. 

Fhe  square  of  either  axis,  is  to  the  square  of  the  other,  as  the 
rectangle  of  the  abscissas  of  the  former,  is  to  the  square  of  their 
ordinate. 

Let  DE  be  an  ordinate  to  the  major  axis  from  the  point 
D  ;  then  we  shall  have 

CA3:CB3::  AExEA':DEa. 

Draw  DTT'  a  tangent  to  the  hyperbola  at  D ;  then,  by 
Prop,  X,  CT  :  CA  :  :  CA  :  CE. 


2J4 


CONIC    SECTIONS. 


Hence  (Prop.  XII.,  B.  II.) 

CA2  :  CE2  :  :  CT  :  CE ; 
and,  by  division  (Prop.  VII.,  B.  II.), 
CA2  :  CE'-CA2 :  :  CT  :  ET.    (1) 
Again,  by  Prop.  X., 

CT'  :  CB  :  :  CB  :  CE'  or  DE. 
Hence  (Prop.  XIL,  B.  II.), 

CBa  :  DE3  :  :  CT'  :  DE. 
But,  by  similar  triangles, 

CT' :  DE   :  :  CT  : 
therefore  CB2  :  DE2  :  :  CT  : 

Comparing  proportions  (1)  and  (2),  we  have 
CA2  :  CE2— CA2  :  :  CB2  :  DE2. 
But  CE2— CA2  is  equal  to  AE  x  EA'  (Prop.  X.,  B.  IV.) ;  hence 

CA2:CB2:  :  AExEA':DEa. 
In  the  same  manner  it  may  proved  that 

CB3  :  C A3  :  :  BE'  X  E'B' :  D'E'2- 
Therefore,  the  square,  &c. 

Cor.  1.          CA2  :  CB2  :  :  CE2— CA2  :  DE2. 
Cor.  2.  The  squares  of  the  ordinates  to  either  axis,  are  to 
each  other  as  the  rectangles  of  their  abscissas. 

Cor.  3.  If  a  circle  be  described  on 
the  major  axis,  then  any  tangent  to 
the  circle,  is  to  the  corresponding  or- 
dinate  in  the  hyperbola,  as  the  major 
axis  is  to  the  minor  axis. 
For,  by  the  Proposition, 

CAa:CB3:  :  AExEA'rDE3. 
But  AE  xEA'  is  equal  to  GE2  (Prop. 
XXVIII.,  B.  IV.). 

Therefore  CA2  :  CB2  :  :  GE2  :  DE2, 

or  CA  :  CB  :  :  GE  :  DE. 


PROPOSITION   XIII.       THEOREM. 

The  lotus  rectum  is  a  third  proportional  to  the  major  ana 
minor  axes. 


Let  LL/  be  a  double  ordinate  to 
*he  major  axis  passing  through  t.ie 
focus  F  ;  then  we  shall  have 

AA'  :  BB'  :  :  BB'  :  LL'. 
Because  LF  is  an  ordinate  to  the  ma- 
jor axis, 

<YC' :  BC2 : :  AF  X  FA' :  LF'  (Prop.  XIL) 
::BC2:  LP  (Prop.  IV.) 


HYPERBOLA. 


215 


Hence  AC  :  BC  :  :  BC  :  LF, 

or  AA'  :  BB'  :  :  BB'  :  LI/. 

Therefore,  the  latus  rectum,  &c. 


PROPOSITION   XIV.       THEOREM. 

If  from  the  vertices  of  two  conjugate  diameters,  ordinates  are 
drawn  to  either  axis,  the  difference  of  their  squares  will  be 
equal  to  the  square  of  half  the  other  axis. 

Let  DD',  EE'  be  any  two  conju- 
gate diameters,  DG  and  EH  ordinates 
to  the  major  axis  drawn  from  their 
vertices,  in  which  case,  CG  and  CH 
will  be  equal  to  the  ordinates  to  the 
minor  axis  drawn  from  the  same 
points  ;  then  we  shall  have 
CA2=CG2-CH3,  and  CB2=EHa-DG2. 

Let  DT  be  a  tangent  to  the  curve  at 
D,  and  ET>  a  tangent  at  E.     Then,  by  Prop.  X., 


whence 

CG 


CGxCT  is  equal  to  CAa,  or  CHxCT' 


CH 


CT'  :  CT  ;  or,  by  similar  triangles, 
:  :  CE   :  DT  ;  that  is, 
:  :  CH  :  GT. 
Hence  CH3=GTxCG 

=  (CG— CT)xCG 
=CG2-CGxCT 

=CG2-CAa(Prop.  X.); 
that  is  CA2=CG2-CH2. 

In  the  same  manner  it  may  be  proved  that 

CB2=EH2-DG2. 
Therefore,  if  from  the  vertices,  &c. 

Cor.  1.  CH2  is  equal  to  CG2-CAa;  that  is,CGxGT;  heice 
(Prop.  XII.,  Cor.  1), 

CA2:CB2::CGxGT:DG2 
Cor.  2.  By  Prop.  X1L, 

CB2  :  CA2  :  :  EH2— CB2  :  CH  . 
By  composition,  " 

CBa    CA2  :  :  EH2  :  CA2-f  CH2  or  CG' 
Hence  CA8 :  CB3 :  :  CG3  :  EH'. 


916 


COMC    SECTIONS. 


PROPOSITION    XV.       THEOREM. 

The  difference  of  the  squares  of  any  two  conjugate  diameters* 
is  equal  to  th&  difference  of  the  squares  of  the  axes. 

Let  DD',  EE'   be  any  two  conju- 
gate diameters  ;  then  we  shall  have 
DD'a— EE'a=AA'a— BB>2. 

Draw  DG,  EH  ordinates  to  the  ma- 
jor axis.  Then,  by  the  preceding 
Proposition, 

CG2— CH2=CAa, 
and  EH2-DG9=CB2. 

Hence         CG2+DG2-CH2— EHa=CA2— CBa, 
or  CD2— CEa=CA2-CB2 ; 

that  is,  DD'a— EE'2=AA'2  — BB'2. 

Therefore,  the  difference  of  the  squares,  &c. 


PROPOSITION    XVI.       THEOREM. 

The  parallelogram  formed  by  drawing  tangents  through  the 
vertices  of  two  conjugate  diameters,  is  equal  to  the  rectangle  of 
the  axes. 

Let  DED'E'  be  a  parallelogram, 
formed  by  drawing  tangents  to  the 
conjugate  hyperbolas  through  the 
vertices  of  two  conjugate  diame- 
ters DD',  EE' ;  its  area  is  equal  to 
AA'XBB'. 

Let  the  tangent  at  D  meet  the 
major  axis  in  T  ;  join  ET,  and.  draw 
the  ordinates  DG,  EH. 

Then,  by  Prop.  XIV.,  Cor.  2,  we  have 


or 
But 

hence 
or 


CAa :  CB3 
CA  :CB 
CT  :CA 
CT  :CB 


:  CG2  :  EH2, 

:  CG  :  EH. 

:  CA  :  CG  (Prop.  X.) ; 

:  CA  :  EH, 


CA  X  CB  is  equal  to  CT  x  EH, 
which  is  equal  to  twice  the  triangle  CTE,  or  the  parallelo- 
gram DE ;   since  the  triangle  and  parallelogram  have  the 
same  base  CE,  and  are  between  the  same  parallels. 

Hence  4CAxCB  or  AA'xBB'   is  equal  to  4DE,  or  the 
uarallelogram  DED'E'.     Therefore,  the  parallelogram,  &c. 


HYPERBOLA. 


217 


PROPOSITION    XVII.       THEOREM. 

If  from  tie  vertex  of  any  diameter,  straight  lines  are  drawn 
to  ike  foci,  their  product  is  equal  to  the  square  of  half  the  con- 
Jugate  diameter. 

Let  DD',  EE'  be  two  conjugate 
diameters,  and  from  D  let  lines 
be  drawn  to  the  foci;  then  will 
FDxF'DbeequaltoEC2. 

Draw  a  tangent  to  the  hyper- 
bola at  D,  and  upon  it  let  fall  the 
perpendiculars  FG,  F'H ;  draw, 
also,  DK  perpendicular  to  EE'. 

Then,  because  the  triangles 
DFG,  DLK,  DF'H  are  similar, 
we  have 

FD:FG :  :DL 

Also,  F'D  :  F'H  :  :  DL 

Whence  (Prop.  XL,  B.  II.), 

FDxF'D:  FGxF'H  ::DL2 
But,  by  Prop.  XVI.,  ACxBC=ECxDK; 
whence  AC  or  DL  :  DK  :  :  EC  :  BC, 

and  DL2  :  DK2  :  :  EC2  :  BC2. 


DK5 


0) 


(2) 


Comparing  proportions  (1)  and  (2),  we  have 

FDxF'D  :  FGxF'H  :  :  EC2  :  BC2. 

But  FGxF'H  is  equal  to  BC2  (Prop.  iX.) ;  hence  FDxF'D 
,s  equal  to  EC2.     Therefore,  if  from  the  vertex,  &c. 


PROPOSITION    XVIII.       THEOREM. 

If  a  tangent  and  ordinate  be  drawn  from  the  same  point  of 
an  hyperbola  to  any  diameter,  half  of  that  diameter  will' be  a 
mean  proportional  between  the  distances  of  the  two  intersections 
from  the  center. 

Let  a  tangent  EG  and  an  ordinate  EH  be  drawn  from  the 
same  point  E  of  an  hyperbola,  meeting  the  diameter  CD 
produced  ;  then  we  shall  have 

CG  :  CD  :  :  CD  :  CH. 

Produce  GE  and  HE  to  meet  the  major  axis  in  K  and  L ; 
drav/  DT  a  tangent  to  the  curve  at  the  point  D,  and  draw 
.DM  A  irallel  to  GK.     Also,  draw  the  ordinates  EN,  DO. 
B,  °rop.  XIV.,  Cor.  1,  CA2  :  CB2  :  :  COxOT  :  DO2, 

:  :CNxNK:EN3. 
K 


CONIC    SECTIONS 


T    L    KANM 
Hence 
"OxOT  :  CNxNK  :  :  DO3  :  EN3 

:  :  OT3  :  NL3,  by  similar  triangles.     (I) 
Also,  by  similar  triangles,  OT  :  NL  :  :  DO  :  EN 

:  :  OM  :  NK.  (2) 

Multiplying  together  proportions  (1)  and  (2)  (Prop.  XL, 
B.  II.),  and  omitting  the  factor  OT3  in  the  antecedents,  and 
NKxNL  in  the  consequents,  we  have 

CO  :  CN  :  :  OM  :  NL ; 

and,  by  division,         CO  :  CN  :  :  CM  :  CL.  (3) 

Also,  by  Prop.  X.,  CKxCN-CA2=CTxCO; 
hence  CO  :  CN  :  :  CK  :  CT.  (4) 

Comparing  proportions  (3)  and  (4),  we  have 

CK  :  CM  :  :  CT  :  CL. 
But  CK  :  CM  :  :  CG  :  CD, 

and  CT  :  CL  :  :  CD  :  CH ; 

hence  CG  :  CD  :  :  CD  :  CH. 

Therefore,  if  a  tangent,  &c. 

Scholium.  The  same  property  may  be  demonstrated  when 
the  tangent  and  ordinate  are  drawn  to  the  conjugate  diameter. 

PROPOSITION    XIX.       THEOREM. 

The  square  of  any  diameter,  is  to  the  square  of  its  conjugate, 
as  the  rectangle  of  its  abscissas,  is  to  the  square  of  their  ordinate. 

Let  DD',  EE'  be  two  conju- 
gate diameters,  and  GH  an  or- 
dinate to  DD' ;  then 
DD'3 :  EE'3 : :  DH  xHD' :  GH2. 

Draw  GTT'  a  tangent  to  the 
curve  at  the  point  G,  and  draw 
GK  an  ordinate  to'EE'.  Then, 
by  Prop.  XVIII., 

CT  :  CD  :  :  CD  :  CH, 
and  CD3  :  CH3  :  :  CT  :  CH 
'1'ron.  XII..  B.  II.), 


HYPERBOLA.  219 

*  whence,  by  division,  CD2  :  CH2  -  CD3  :  :  CT  :  IIT.        (1) 
Also,  by  Prop.  XVIIL,  Scholium,  CT'  :  CE  ::  CE  :  CK, 
and  CEa  :  CK3  :  :  CT'  :  CK  or  GH, 

:  :  CT  :  HT.  (2) 

Comparing  proportions  (1)  and  (2),  we  have 

CD2  :  CE2  :  :  CH2-CD3  :  CK3  or  GH3, 
or  DD>3 :  EE'2 :  :  DH  x  HD'  :  GH3. 

Therefore,  the  square,  &c. 

Cor.  1.  In  the  same  manner  it  may  be  proved  that  DD'* . 
EE'3  :  :  DHxHD'  :  G'H2;  hence  GH  is  equal  to  G'H,  or 
every  diameter  bisects  its  double  ordinates. 

Cor.  2.  The  squares  of  the  ordinates  to  any  diameter,  ar«» 
to  each  other  as  the  rectangles  of  their  abscissas. 


PROPOSITION   XX.       THEOREM. 

If  a  cone  be  cut  by  a  plane,  not  passing  through  the  vertex^ 
and  making  an  angle  with  the  base  greater  than  that  made  by 
the  side  of  the  cone,  the  section  is  an  hyperbola. 

Let  ABC  be  a  cone  cut  by  a  plane 
DGH,  not  passing  through  the  vertex, 
and  making  an  angle  with  the  base 
greater  than  that  made  by  the  side  of 
the  cone,  the  section  DHG  is  an  hyper- 
bola. 

Let  ABC  be  a  section  through  the  axis 
of  the   cone,  and   perpendicular  to  the 
plane    HDG.      Let    bgcd   be    a    section 
made  by  a  plane  parallel  to  the  base  of  ^< 
the  cone ;   then  DE,  the  intersection  of 
the  planes  HDG,  BGCD,  will  be  perpen- 
dicular to  the  plane  ABC,  and,  consequently,  to  each  of  the 
lines  BC,  HE.     So,  also,  de  will  be  perpendicular  to  be  and 
HE.     Let  AB  and  HE  be  produced  to  meet  in  L. 

Now,  because  the  triangles  LBE,  Ue  are  similar,  as  also 
the  triangles  HEC,  Hec,  we  have  the  proportions 
BE  :  be  :  :  EL  :  eL 
EC  :  ec  :  :  HE  :  He. 
Hence,  by  Prop.  XL,  B.  II., 

BExEC  :  bey.ec  :  :  HExEL  :  HeXeL. 
But,  since  BC  is  a  diameter  of  the  circle  BGCD,  and  DE  is 
perpendicular  to  BC,  we  have  (Prop.  XXIL,  Cor.,  B.  IV.), 
BExEC  =  DE3. 

For  the  same  reason, 


820 


CONIC    SECTIONS 


Substituting  these  values  of  BE  xEC  and  beXec,  in  the  pre-* 
ceding  proportion,  we  have 

DE2:de2:  :  HExEL:HexcL; 

that  is,  the  squares  of  the  ordinates  to  the  diameter  HE,  are 
to  each  other  as  the  products  of  the  corresponding  abscissas. 
Therefore  the  curve  is  an  hyperbola  (Prop.  XII.,  Cor.  2) 
whose  major  axis  is  LH.  Hence  the  hyperbola  is  called  a 
a  aic  section,  as  mentioned  on  page  177 


OF  THE  ASYMPTOTES. 


Definition.  —  An  asyjnptote  of  an  hyperbola  is  a  straight 
iins  drawn  through  the  center,  which  approaches  nearer  the 
curve,  the  further  it  is  produced,  but  being  extended  ever  so 
far,  can  never  meet  the  curve. 


PROPOSITION    XXI.       THEOREM. 

If  tangents  to  four  conjugate  hyperbolas  be  drawn  througfi 
the  vertices  of  the  axes,  the  diagonals  of  the  rectangle  so  formed 
ire  asymptotes  to  the  curves. 

Let  AA',  BB'  be  the  axes  of 
four  conjugate  hyperbolas,  and 
through  the  vertices  A.  A',  B, 
B',  let  tangents  to  the  curve  be 
drawn,  and  let  CE,  CE'  be  the 
diagonals  of  the  rectangle  thus 
:ied;  CE  and  CE'  will  be 
asymptotes  to  the  curves. 

From  any  point  D  of  one  of  the 
curves,   draw   the   ordinate   DG, 
and  produce  it  to  meet  CE  in  H. 
Then,  from  similai;  triangles,  we  shall  have 
CG2  :  GH2  :  :  CA2  :  AE2  or  CB2, 

:  :  CG'-CA'  :  DG2  (Prop.  XII.,  Cor.  1). 

Now,  according  as  the  ordinate  DG  is  drawn  at  a  greater 
distance  from  the  vertex,  CG3  increases  in  comparison  with 
CA8;  that  is,  the  ratio  of  CG2  to  CG2  — CA2  continually  ap- 
proaches to  a  ratio  of  equality.  But  however  much  CG  may 
be  increased,  GG2  —  CA2  can  never  become  equal  to  CG2 : 
hence  DG  can  never  become  equal  to  HG,  but  approaches 
continually  nearer  to  an  equality  with  it,  the  further  we  re- 
cede from  the  vertex.  Hence  CH  is  an  asymptote  of  the 
hyper)  o!n ;  since  it  is  a  line  drawn  through  the  center,  which 


HYPERBOLA. 


approaches  nearer  the  curve,  the  further  it  is  pioduced.  but 
being  extended  ever  so  far,  can  never  meet  the  curve. 

In  the  same  manner  it  may  be  proved  that  CH'  is  an 
asymptote  of  the  conjugate  hyperbola. 

Co; .  1.  The  two  asymptotes  make  equal  angles  with  the 
major  axis,  and  also  with  the  minor  axis. 

Cor.  2.  The  line  AB  joining  the  vertices  of  the  two  axes,  is 
bisected  by  one  asymptote,  and  is  parallel  to  the  other. 

Cor.  3.  All  lines  perpendicular  to  either  axis,  and  termi- 
nated by  the  asymptotes,  are  bisected  by  that  &xis 


PROPOSITION   XXII.       THEOREM. 

If  an  ordinate  to  either  axis  be  produced  to  meet  the  asymp 
totes,  the  rectangle  of  the  segments  into  which  it  is  divided  by 
the  curve,  will  be  equal  to  the  square  of  half  the  other  axis. 

Let  DG  be  an  ordinate  to  the 
major  axis,  and  let  it  be  produced 
to  meet  the  asymptotes  in  II  and 
H' ;  then  will  the  rectangle  HD  X 
DH'  be  equal  to  BC2. 

For,  by  Prop.  XII.,  Cor.  1, 
CA2  :  AE2 :  :  CG2-CA2 :  DG2 , 
or,  by  similar  triangles, 

:  :  CG2  :  GH2. 
flence 

CG2  :  GH2  :  :  CG2_CA2 :  DG2, 
and,  by  division, 

CG2  :  GH2  :  :  CA2  :  GH'-DG2,  or  as  CA2  :  AE2. 
Since  the  antecedents  of  this  proportion  are  equal  to  each 
other,  the  consequents  must  be  equal  ;  that  is, 

AE2  or  BC2  is  equal  to  GH2— DG2 ; 
which  is  equal  to  HD  xDH'. 

So,  also,  it  may  be  proved  that 

CA2=D'KXD'L. 

Cor.  HDxDH'=BCa  =  KMxMK';  that  is,  if  ordinates  to 
the  major  axis  be  produced  to  meet  the  asymptotes,  the  rect- 
angles of  the  segments  into  which  these  lines  are  divided  by 
the  curve  are  equal  to  each  other. 


222 


CONIC  SECTIONS. 


PROPOSITION   XXIII.       THEOREM. 

All  the  $  arallelograms  formed  by  drawing  lines  from  any 
point  of  an  hyperbola  parallel  to  the  asymptotes,  are  equal  ic 
each  other. 

Let  CH,  CH'  be  the  asymptotes 
of  an  hyperbola ;  let  the  lines  AK, 
DL  be  drawn  parallel  to  CH',  and 
the  lines  AK',  DL'  parallel  to  CH ; 
then  will  the  parallelogram  CLDL' 
be  equal  to  the  parallelogram 
CKAK'. 

Through  the  points  A  and  D 
draw  EE',  HH',  perpendicular  to 
the  major  axis ;  then,  because  the 
triangles  AEK,  DHL  are  similar, 
as  also  the  triangles  AE'K',  DH'L', 
we  have  the  proportions 

AK  :  AE  :  :  DL  :  DH. 
Also,  AK'  :  AE'  :  :  DL'  :  DH'. 
Hence  (Prop.  XL,  B.  II), 

AKxAK'  :  AExAE'  :  :  DLxDL'  :  DHxDH'. 
But,  by  Prop.  XXII,  the  consequents  of  this  proportion  are 
equal  to  each  other;  hence 

AKxAK'  is  equal  to  DLxDL'. 

But  the  parallelograms  CA,  CD  being  equiangular,  are  as 
the  rectangles  of  the  sides  which  contain  the  equal  angles 
(Prop.  XXIII,  Cor.  2,  B.  IV.)  ;  hence  the  parallelogram  CD 
is  equal  to  the  parallelogram  CA. 

Cor.  Because  the  area  of  the  rectangle  DL  x  DL'  is  con 
stant,  DL  varies  inversely  as  DL' ;  that  is,  as  DL'  increases, 
DL  diminishes  ;  hence  the  asymptote  continually  approaches 
the  curve,  but  never  meets  it.  The  asymptote  CH  may, 
therefore,  be  considered  as  a  tangent  to  the  curve  at  a  point 
infinitely  distant  from  C. 


NOTES. 


I  IGE  9,  Def.  III. — For  the  sake  of  brevity,  the  word  line  is  often  used  to  des 
Igm  le  a  straight  line. 

P.  12,  Ax.  II. — This  axiom,  when  applied  to  geometrical  magnitudes;  must  be 
undiTstood  to  refer  simply  to  equality  of  areas.  It  is  not  designed  to  assert  that, 
whe.i  equal  triangles  are  united  to  equal  triangles,  the  resulting  figures  wih 
idmii  of  coincidence  by  superposition. 

P.  32,  Prop.  XXVIII. — When  this  proposition  is  applied 
to  polygons  which  have  re-entering  angles,  each  of  these  an- 
gles ia  to  be  regarded  as  greater  than  two  right  angles.  But, 
«n  order  to  avoid  ambiguity,  we  shall  confine  our  reasoning 
to  polygons  which  have  only  salient  angles,  and  which  may 
be  called  convex  poly  gong.  Every  convex  polygon  is  such, 
that  a  straight  line,  however  drawn,  can  not  meet  the  pe- 
rimeter of  the  polygon  in  more  than  two  points. 

P.  32,  Cor.  2. — This  corollary  supposes  that  all  the  sides  of  the  polygon  are 
produced  outward  in  the  same  direction. 

P.  53,  Props.  XII.  and  XIII. — It  will  be  perceived  that  the  relative  situation 
of  two  circles  may  present  five  cases. 

1st.  When  the  distance  between  their  centers  is  greater  than  the  sum  of  their 
radii,  there  can  be  neither  contact  nor  intersection. 

2d.  When  the  distance  between  then:  centers  is  equal  to  the  sum  of  their 
radii,  there  is  an  external  contact. 

3d.  When  the  distance  between  their  centers  is  less  than  the  sum  of  their 
radii,  but  greater  than  their  difference,  there  is  an  intersection. 

4th.  When  the  distance  between  their  centers  is  equal  to  the  difference  of 
their  radii,  there  is  an  internal  contact. 

5th.  When  the  distance  between  their  centers  is  less  than  the  difference  of 
their  radii,  there  can  be  neither  contact  nor  intersection. 

P.  55,  Cor.  1. — An  angle  inscribed  in  a  segment  is  the  angle  contained  by  two 
straight  lines  drawn  from  any  point  in  the  circumference  of  the  segment  to  the 
extremities  of  the  chord,  which  is  the  base  of  the  segment. 

P.  63,  Prop.  VIII. — Every  right-angled  parallelogram,  or  rectangle,  is  said  to  be 
coiitained  by  any  two  of  the  straight  lines  which  are  about  one  of  the  right  angles 

P.  70,  Scholium. — By  the  segments  of  a  line  we  understand  the  portions  into 
which  the  line  is  divided  at  a  given  point.  So,  also,  by  the  segments  of  a  line 
produced  to  a  given  point,  we  are  to  understand  the  distances  between  -the  giv 
en  point  and  the  extremities  of  the  line. 

P.  71,  Props.  XVIII.  and  XIX. — It  will  be  perceived  by  these  two  propositions, 
that  when  the  angles  of  one  triangle  are  respectively  equal  to  those  of  another, 
the  sides  of  the  former  are  proportional  to  those  of  the  latter,  and  conversely ; 
so  that  either  of  these  conditions  is  sufficient  to  determine  the  similarity  of  two 
triangles.^  This  is  not  true  of  figures  having  more  than  three  sides ;  for  with  re 
spect  to  these  of  only  four  sides,  or  quadrilaterals,  we  may 
alter  the  proportion  of  the  sides  without  changing  the 
angles,  or  change  the  angles  without  altering  the  sides ; 
thus,  because  the  angles  are  equal,  it  does  not  follow 
that  the  sides  are  proportional,  or  the  converse.  It  is 
evident,  for  example,  that  by  drawing  EF  parallel  to 
BC,  the  angles  of  the  quadrilateral  AEFD  are  equal  to 
those  of  the  quadrilateral  ABCD,  but  the  proportion  of 
the  sides  is  different.  Also,  without  changing  the  four 
sides  AB,  BC,  CD,  DA,  we  can  make  the  point  A  ap-  A 
proach  C,  or  recede  from  it,  wvrich  would  change  the  angles. 

These  two  propositions,  which,  properly  speaking,  form  but  one,  together 
with  Prop.  XL,  are  the  most  important  and  the  most  fruitful  in  results  of  any  in 
Geometry.  They  are  almost  sufficient  of  themselves  for  all  subsequent  applica- 

-3,  ani  for  the  resolution  of  every  problem.     The  reason  is,  that  all  figures 


NOTES. 


may  be  divided  into  triangles,  and  any  triangle  into  two  right-angled  triangles 
Thus,  the  general  properties  of  triangles  involve  those  of  all  rectilineal  figures. 

Page  113,  Prop.  II. — In  this  and  the  following  propositions,  the  planes  spokea 
of  are  supposed  to  be  of  indefinite  extent. 

P.  157,  Prop.  X In  all  the  preceding  propositions  it  has  been  supposed,  in 

conformity  with  Def.  6,  that  spherical  triangles  always  have  each  of  their  sidea 
less  than  "a  semicircumfereuce ;  in  which  case  their  angles  are  always  less  than 
two  right  angles.  For  if  the  side  AB  is  less  than 
a  semicircumference,  as  also  AC,  both  of  these 
arcs  must  be  produced,  in  order  to  meet  in  D. 
Now  the  two  angles  ABC,  DBC,  taken  together, 
are  equal  to  two  right  angles ;  therefore  the  angle 
ABC  is  by  itself  less  than  two  right  angles. 

It  should,  however,  be  remarked  that  there 
are  spherical  triangles,  of  which  certain  sides  are 
greater  than  a  semicircumference,  and  certain  an- 
gles greater  than  two  right  angles.  For  if  we 
produce  the  side  AC  so  as  to  form  an  entire  cir- 
cumference, ACDE,  the  part  which  remains,  after 
taking  from  the  surface  of  the  hemisphere  the  triangle  ABC,  is  a  new  triangle, 
which  may  also  be  designated  by  ABC,  and  the  sides  of  which  are  AB,  BO, 
CDEA.  Here  we  see  that  the  side  CDEA  is  greater  than  the  semicircumfer- 
ence DEA,  and  at  the  same  time  the  opposite  angle  ABC  exceeds  two  right 
angles  by  the  quantity  CBD. 

Triangles  whose  sides  and  angles  are  so  large  have  been  excluded  by  the 
definition,  because  their  solution  always  reduces  itself  to  that  of  triangles  em- 
braced in  the  definition.  Thus,  if  we  know  the  sides  and  angles  of  the  triangle 
ABC,  we  shall  know  immediately  the  sides  and  angles  of  the  trian-gle  of  the 
same  name,  which  is  the  remainder  of  the  surface  of  the  hemisphere. 

P.  178. — The  subtangent  is  so  called  because  it  is  below  the  tangent,  being 
limited  by  the  tangent  and  ordinate  to  the  point  of  contact.  The  subnormal  is 
so  called  because  it  is  below  the  normal,  being  limited  by  the  normal  and  crdi- 
nate.  The  subtangent  and  subnormal  may  be  regarded  as  the  projections  cf  tl*i 
tangent  and  normal  upon  a  diameter. 


P.  179,  Prop.  I. — By  the  method  here  indicated  a 
parabola  may  be  described  with  a  continuous  motion. 
It  may,  however,  be  described  by  points  as  follows: 

In  the  axis  produced  take  VA  equal  to  VF,  the  focal 
distance,  and  draw  any  number  of  Hues,  BB,  B'B' 
etc.,  perpendicular  to  the  axis  AD;  then,  with  the 
distances  AC,  AC',  AC",  etc.,  as  radii,  and  the  focus 
F  as  a  center,  describe  arcs  intersecting  the  perpen- 
diculars in  B,  B',  etc.  Then,  with  a  steady  hand, 
draw  the  curve  through  all  the  points  B,  B',  B",  etc. 


r.' 


P.  179,  Prop.  II.— It  may  be  thought  that  if  the 
point  D  can  not  lie  on  the  curve,  it  may  fall  within 
it,  as  is  represented  in  the  annexed  figure.  This 
may  be  proved  to  be  impossible,  as  follows : 

Let  the  line  DE,  perpendicular  to  the  directrix, 
meet  the  curve  in  G,  and  ioin  FG.  Now,  by  Prop. 
VIII.,  B.  L, 

FG+GD>FD. 

Hence  FG>FD— GD, 

>ED-GD, 
That  ia,  FG  is  greater  than  EG  which  is  contrary  to 


NOTES. 


225 


Page  183,  Prop.  VIII. — As  no  attempt  is  here  made  to  compare  figures  by  RU 
•perposition,  the  equality  spoken  of  is  only  to  be  understood  as  implying  equal 
areas.  Throughout  the  remainder  of  this  treatise  the  word  equal  is  employed 
instead  of  equivalent. 

P.  185,  Prop.  XL — The  conclusion  that  DVG  is  a  parabola  would  not  be 
legitimate,  unless  it  was  proved  that  the  property  that  "  the  squares  of  the  ordi 
Dates  are  to  each  other  as  the  corresponding  abscissas" 
is  peculiar  to  the  parabola.  That  such  is  the  case,  ap- 
pears from  the  fact  that,  when  the  axis  and  one  point 
of  a  parabola  are  given,  this  property  will  determine 
the  position  of  every  other  point.  Thus,  let  VE  be 
the  axis  of  a  parabola,  and  g  any  point  of  the  curve, 
from  which  draw  the  ordinate  ge.  Take  any  other 
point  in  the  axis,  as  E,  and  make  GE  of  such  a  length 
that  Ve  :  VE  :  :  ge*  :  GE2. 

Since  the  first  three  terms  of  this  proportion  are  given,  the  fourth  is  de- 
termined,  and  the  same  proportion  will  determine  any  number  of  points  cf  the 
curve. 

A  similar  remark  is  applicable  to  Prop.  XX.  of  the  Ellipse  and  Hyperbola. 

P.  196,  Prop.  X. — It  may  be  proved  that  CT'  :  CB  : :  CB  :  CG'  in  the  follow 
ing  manner.  Draw  DH  perpendic-  m/ 

ular  to  TT',  and  it  will  bisect  the 
angle  FDF'. 

Hence 

F'H  :  HF  :  :  F'D  :  DF, 
:  :  F'T  :  FT. 
Therefore,  Prop.  VII.,  Cor.  B.  II.. 

2CF  :  2CH  :  :  2CT  :  2CF. 
Whence   CTxCH^CF*. 
But  we  have  proved  that 
CTxCG=CA3. 
Hence  CTxGH=CA2— CF2=CB2. 

Again,  because  the  triangles  CTT'  and  DGH  are  similar,  we  have 

CT  :  CT' :  :  DG  :  GH. 

Whence  CTxGH=CT'xDG=:CT'xCG'; 

Therefore,  CT'xCG'^CB2, 

or  CT'  :  CB  :  :  CB  :  CG'. 

The  following  demonstration  of  Prop.  X.  was  suggested  to  me  by  Professor 
J.  H.  Coffin. 

Let  TT'  be  a  tangent  to  the  ellipse,  and  DG  an  ordinate  to  the  major  axis  from 


the  point  of  contact ;  then  we  shall  have 

CT  :  CA  :  :  CA  :  CG. 

From  F  draw  FH  perpendicular  to  TT',  and  join  DF,  DF',  CH,  and  GH.  Then, 
by  Prop.  VIII.,  Cor.,  CH  is  parallel  to  DF' ;  and  since  DGF,  DHF  are  both  right 
angles,  a  circle  described  on  DF  as  a  diameter  will  pass  through  the  points  G 
and  H.  Therefore,  the  angle  HGF  is  equal  to  the  angle  HDF  (Prop.  XV.,  Cor.  1, 
B.  Ill),  which  is  equal  to  T'DF'  or  DHC.  Hence  the  angles  CGH  and  CHT 
which  are  the  supplements  of  HGF  and  DHC,  are  equal.  And  since  the  angle 
C  is  common  to  the  two  triangles  CGH,  CHT,  they  are  equiangular,  and  w$ 
have  CT  :  CH  :  :  CH  :  CG. 

But  CH  is  equal  to  CA  (Prop.  VIII);  therefore 

CT  :  CA  :  :  CA  :  CG. 


226 


NOTES. 


Page  .98,  Prop.  XIV.— That  the  triangles  CDT,  GET'  aie  similar,  may  b* 
proved  as  follows  • 

AG.GA'=CAa-CG2 

=CG.CT-CG2,  Prop.  X. 

=CG.GT.  /i) 

In  the  same  manner,  AH.HA'=CH.HT'. 
Since  the  triangles  DGT,  EHC  are  similar, 

GT  :  CH  :  :  DG  :  EH  ; 
or  GT3 :  CH2 : :  DG8 :  EH2 ; 

::AG.GA':AH.HA.     Prop.  XII.,  Cor.  2 
:  :  CG.GT  :  CH.HT ,  by  Equation  (1), 
Therefore,         CG  :  HT' :  :  GT  :  CH 
: :  DG  :  EH. 

Hence  the  triangles  CDG,  EHT'  are  similar ;  and,  therefore,  the  whole  triangles 
CDT,  CET'  are  similar. 

Page  207,  Prop.  I. — The  hyperbola  may  be 
described  by  points,  as  follows : 

In  the  major  axis  AA'  produced,  take  the  foci 
F,  F'  and  any  point  D.  Then,  with  the  radii  AD, 
A'D,  and  centers  F,  F',  describe  arcs  intersecting 
each  other  in  E,  which  will  be  a  point  in  the 
curve.  In  like  manner,  assuming  other  points, 
D',  D",  etc.,  any  number  of  points  of  the  curve 
may  be  found.  Then,  with  a  steady  hand,  draw 
the  curve  through  all  the  points  E,  E',  E",  etc. 

In  the  same  manner  may  be  constructed  the 
two  conjugate  hyperbolas,  employing  the  axis  BP' 

P.  209,  Prop.  V — It  may  be  thought 
that  if  the  point  E  can  not  lie  on  the 
curve,  it  may  fall  within  it,  as  is  repre- 
sented in  the  annexed  figure.  This  may 
be  proved  to  be  impossible,  as  follows : 

Join  EF',  meeting  the  curve  in  K,  and 
Join  KF.  Now,  by  Prop.  VIII.,  B.  I., 

FK>EF-EK; 
therefore, 

F'K-FK<F'K+EK-EF 

<EF'-EF; 

But       EF'— EF=F'G=DF/— DF. 
Hence       F'K-FK<DF'-DF, 
which  is  contrary  to  Def.  1. 

P.  212,  Prop.  X. — This  proposition  may  be  otherwise  demonstrated,  lik* 
Prop  X.  af  the  Ellipse. 


EDtfD* 


GEOMETRICAL  EXERCISES. 


A  FEW  theorems  without  demonstrations,  and  problems 
without  solutions,  are  here  subjoined  for  the  exercise  of  the 
pupil.  They  will  be  found  admirably  adapted  to  familiarize 
the  beginner  with  the  preceding  principles,  and  to  impart  dex- 
terity in  their  application.  No  general  rules  can  be  prescribed 
which  will  be  found  applicable  in  all  cases,  and  infallibly  lead 
to  the  demonstration  of  a  proposed  theorem,  or  Hhe  Solution 
of  a  problem,  The  following  directions  may  prove  of  some 
service. 

ANALYSIS  OF  THEOREMS. 

1.  Construct  a  diagram  as  directed  in  the  enunciation,  and 
assume  that  the  theorem  is  true. 

2.  Consider  what  consequences  result  from  this  admission, 
by  combining  with  it  theorems  which  have  been  already 
proved,  and  which  are  applicable  to  the  diagram. 

3.  Examine  whether  any  of  these  consequences  are  already 
known  to  be  true  or  to  be  false, 

4.  If  any  one  of  them  be  false,  we  have  arrived  at  a  reduc- 
tio  ad  absurdum,  which  proves  that  the  theorem  itself  is  false, 
as  in  Book  I.-,  Prop.  4,  16,  etc. 

5.  If  none  of  the  consequences  so  deduced  be  known  to  be 
either  true  or  false,  proceed  to  deduce  other  consequences 
from  all  or  any  of  these  until  a  result  is  obtained  which  is 
known  to  be  either  true  or  false. 

6.  If  we  thus  arrive  at  some  truth  which  has  been  previous- 
ly demonstrated,  we  then  retrace  the  steps  of  the  investiga 
tion  pursued  in  the  analysis,  till  they  terminate  in  the  theorem 
which  was  assumed.     This  process  will  constitute  the  demon- 
stration of  the  theorem. 

ANALYSIS  OF  PROBLEMS. 

1.  Construct  the  diagram  as  directed  in  the  enunciation, 
and  suppose  the  solution  of  the  problem  effected. 

2.  Examine  the  relations  of  the  lines,  angles,  triangles,  etc., 
in  the  diagram,  and  find  the  dependence  of  the  assumed  solu- 
tion on  some  theorem  or  problem  in  the  Geometry. 


228  GEOMETRICAL    EXERCISES.   . 

3.  If  such  can  not  be  found,  draw  other  lines,  parallel  or 
perpendicular,  as  the  case  may  require  ;  join  given  points  or 
points  assumed  in  the  solution,  and  describe  circles  if  neces- 
sary ;  and  then  proceed  to  trace  the  dependence  of  the  as- 
sumed solution  on  some  theorem  or  problem  in  Geometry. 

4.  If  we  thus  arrive  at  some  previously  demonstrated  or  ad 
mitted  truth,  we  shall  obtain  a  direct  solution  of  the  problem 
by  assuming  the  last  consequence  of  the  analysis  as  the  first 
step  of  the  process,  and  proceeding  in  a  contrary  order  through 
the  several  steps  of  the  analysis,  until  the  process  terminate  in 
the  problem  required. 

It  may  perhaps  be  expedient  to  defer  attempting  the  solu- 
tion of  the  following  problems,  until  Book  V.  has  been  studied 


GEOMETRICAL  EXERCISES  ON  BOOK  I. 
THEOREMS. 

Prop.  1 .  The  difference  betwe.en  any  two  sides  of  a  trian- 
gle is  less  than  the  third  side. 

Prop,  2.  The  sum  of  the  diagonals  of  a  quadrilateral  is  less 
than  the  sum  of  any  four  lines  that  can  be  drawn  from  any 
point  whatever  (except  the  intersection  of  the  diagonals)  to 
the  four  angles. 

Prop.  3,  If  a  straight  line  which  bisects  the  vertical  an- 
gle of  a  triangle  also  biseqts  the  base,  the  remaining  sides 
of  the  triangle  are  equal  to  each  other. 

Prop.  4-  If  the  base  of  an  isosceles  triangle  be  produced, 
twice  the  exterior  angle  is  greater  than  two  right  angles  by 
the  vertical  angle. 

Prop.  5.  In  any  right-angled  triangle,  the  middle  point  of 
the  hypothenuse  is  equally  distant  from  the  three  angles. 

Prop.  6.  If  on  the  sides  of  a  square,  at  equal  distances 
from  the  four  angles,  four  points  be  taken,  one  on  each 
side,  the  figure  formed  by  joining  those  points  will  also  be  a 
square. 

Prop.  7.  If  one  angle  of  a  parallelogram  be  a  right  angle, 
the  parallelogram  will  be  a  rectangle. 

Prop.  8.  If  the  diagonals  of  a  quadrilateral  bisect  each  oth- 
er, the  figure  is  a  parallelogram. 

Prop.  9.  The  parallelogram  whose  diagonals  are  equal  is 
rectangular. 

Prop.  10.  Any  line  drawn  through  the  centre  of  the  diag- 
onal of  a  parallelogram  to  meet  the  sides,  is  bisected  in  that 
point,  and  also  bisects  the  parallelogram. 


GEOMETRICAL    EXERCISES.  22k 


PROBLEMS. 

Prop.  1.  On  a  given  line  describe  an  isosceles  triangle,  each 
of  whose  equal  sides  shall  be  double  of  the  base. 

Prop.  2.  On  a  given  line  describe  a  square,  of  which  the 
line  shall  be  the  diagonal. 

Prop.  3.  Divide  a  right  angle  into  three  equal  angles. 

Prop.  4.  One  of  the  acute  angles  of  a  right-angled  triangle 
is  three  times  as  great  as  the  other ;  trisect  the  smaller  of 
these. 

Prop.  5.  Construct  an  equilateral  triangle,  having  given 
the  length  of  the  perpendicular  drawn  from  one  of  the  angles 
on  the  opposite  side. 

GEOMETRICAL  EXERCISES  ON  BOOK  m. 

THEOREMS. 

Prop.  1 .  Every  chord  of  a  circle  is  less  than  the  diameter. 

Prop.  2.  Any  two  chords  of  a  circle  which  cut  a  diameter 
in  the  same  point,  and  at  equal  angles,  are  equal  to  each 
other. 

Prop.  3.  The  straight  lines  joining  toward  the  same  parts, 
the  extremities  of  any  two  chords  in  a  circle  equally  distant 
from  the  centre,  are  parallel  to  each  other. 

Prop.  4.  The  two  right  lines  which  join  the  opposite  ex- 
tremities of  two  parallel  chords,  intersect  in  a  point  in  that 
diameter  which  is  perpendicular  to  the  chords. 

Prop.  5.  All  the  equal  chords  in  a  circle  may  be  touched 
by  another  circle. 

Prop.  6.  The  lines  bisecting  at  right  angles  the  sidles  of  a 
triangle,  all  meet  in  one  point. 

Prop.  7.  If  two  opposite  sides  of  a  quadrilateral  figure  in- 
scribed in  a  circle  are  equal,  the  other  two  sides  will  be  par- 
allel. 

Prop.  8.  If  an  arc  of  a  circle  be  divided  into  three  equal 
parts  by  three  straight  lines  drawn  from  one  extremity  of  the 
arc,  the  angle  contained  by  two  of  the  straight  lines  will  be 
bisected  by  the  third. 

Prop.  9.  If  the  diameter  of  a  circle  be  one  of  the  equal  sides 
of  an  isosceles  triangle,  the  base  will  be  bisected  by  the  cir- 
cumference. 

Prop.  10.  If  two  circles  touch  each  other  externally,  and 
parallel  diameters  be  drawn,  the  straight  line  joining  the  op- 
posite extremities  of  these  diameters  will  pass  through  the 
point  of  contact. 


230  GEOMETRICAL    EXERCISES. 

Prop.  11.  The  lines  which  bisect  the  angles  of  any  paraL 
lelogram  form  a  rectangular  parallelogram,  whose  diagonals 
are  parallel  to  the  sides  of  the  former. 

Prop.  12.  If  two  opposite  sides  of  a  parallelogram  be  bi- 
sected, the  lines  drawn  from  the  points  of  bisection  to  the  op- 
posite angles  will  trisect  the  diagonal. 

PROBLEMS. 

Prop.  1 .  From  a  given  point  without  a  given  straight  line, 
draw  a  line  making  a  given  angle  with  it. 

Prop.  2.  Through  a  given  point  within  a  circle,  draw  a 
chord  which  shall  be  bisected  in  that  point. 

Prop.  3.  Through  a  given  point  within  a  circle,  draw  the 
least  possible  chord. 

Prop,  4.  Two  chords  of  a  circle  being  given  in  magnitude 
and  position,  describe  the  circle. 

Prop.  5.  Describe  three  equal  circles  touching  one  anoth- 
er ;  and  also  describe  another  circle  which  shall  touch  them 
all  three. 

Prop.  6.  How  many  equal  circles  can  be  described  around 
another  circle  of  the  same  magnitude,  touching  it  and  one  an- 
other ? 

Prop.  7.  With  a  given  radius,  describe  a  circle  which  shall 
pass  through  two  given  points. 

Prop.  8.  Describe  a  circle  which  shall  pass  through  two 
given  points,  and  have  its  centre  in  a  given  line. 

Prop.  9.  In  a  given  circle,  inscribe  a  triangle  equiangular  to 
a  given  triangle. 

Prop.  10.  From  one  extremity  of  a  line  which  can  not  be 
produced,  draw  a  line  perpendicular  to  it. 

Prop.  11.  Divide  a  circle  into  two  parts  such  that  the  an- 
gle contained  in  one  segment  shall  equal  twice  the  angle  con- 
tained in  the  other. 

Prop.  12.  Divide  a  circle  into  two  segments  such  that  the 
angle  contained  in  one  of  them  shall  be  five  times  the  angle 
contained  in  the  other. 

Prop.  13.  Describe  a  circle  which  shall  touch  a  given  cir- 
cle in  a  given  point,  and  also  touch  a  given  straight  line. 

Prop.  14.  With  a  given  radius,  describe  a  circle  which  shall 
pass  through  a  given  point  and  touch  a  given  line. 

Prop.  15.  With  a  given  radius,  describe  a  circle  which 
shall  touch  a  gwen  line,  and  have  its  centre  in  another  given 
line. 


GEOMETRICAL    EXERCISES.  231 

GEOMETKICAL  EXERCISES  ON  BOOK  IV. 
THEOREMS. 

Prop.  1.  The  area  of  a  triangle  is  equal  to  its  perimeter 
multiplied  by  half  the  radius  of  the  inscribed  circle. 

Prop.  2.  If  from  any  point  in  the  diagonal  of  a  parallelo- 
gram, lines  be  drawn  to  the  angles,  the  parallelogram  will  be 
divided  into  two  pairs  of  equal  triangles. 

Prop.  3.  If  the  sides  of  any  quadrilateral  be  bisected,  and 
the  points  of  bisection  joined,  the  included  figure  will  be  a 
parallelogram,  and  equal  in  area  to  half  the  original  figure. 

Prop.  4.  Show  how  the  squares  in  Prop.  XL,  Book  IV., 
may  be  dissected,  so  that  the  truth  of  the  proposition  may  be 
made  to  appear  by  superposition  of  the  parts. 

Prop.  5.  In  the  figure  to  Prop.  XL,  Book  IV., 

(a.)  If  BG  and  CH  be  joined,  those  lines  will  be  parallel. 
(b.)  If  perpendiculars  be  let  fall  from  F  and  I  on  BC  pro- 
duced, the  parts  produced  will  be  equal,  and  the  perpendic- 
ulars together  will  be  equal  to  BC. 

(c.)  Join  GH,  IE,  and  FD,  and  prove  that  each  of  the 

triangles  so  formed  is  equivalent  to  the  given  triangle  ABC. 
(d.)  The  sum  of  the  squares  of  GH,  IE,  and  FD  will  be 

equal  to  six  times  the  square  of  the  hypothenuse. 

Prop.  6.  The  square  on  the  base  of  an  isosceles  triangle 
whose  vertical  angle  is  a  right  angle,  is  equal  to  four  times  the 
area  of  the  triangle. 

Prop.  7.  If  from  one  of  the  acute  angles  of  a  right-angled 
triangle,  a  straight  line  be  drawn  bisecting  the  opposite  side, 
the  square  upon  that  line  will  be  less  than  the  square  upon 
the  hypothenuse,  by  three  times  the  square  upon  half  the  line 
bisected. 

Prop.  8.  In  a  right-angled  triangle,  the  square  on  either  of 
the  two  sides  containing  the  right  angle,  is  equal  to  the  rect- 
angle contained  by  the  sum  and  difference  of  the  other  sides. 

Prop.  9.  In  any  triangle,  if  a  perpendicular  be  drawn  from 
the  vertex  to  the  base,  the  difference  of  the  squares  upon  the 
sides  is  equal  to  the  difference  of  the  squares  upon  the  seg- 
ments of  the  base. 

Prop.  10.  The  squares  of  the  diagonals  of  any  quadrilateral 
figure  are  together  double  the  squares  of  the  two  lines  joining 
the  middle  points  of  the  opposite  sides. 

Prop.  11.  If  one  side  of  a  right-angled  triangle  is  double 
the  other,  the  perpendicular  from  the  vertex  upon  the  hypoth- 
enuse will  divide  the  hypothenuse  into  parts  which  are  in  the 
ratio  of  1  to  4. 


232  GEOMETRICAL    EXERCISES. 

Prop.  12.  If  two  circles  intersect,  the  common  chord  pro- 
duced will  bisect  the  common  tangent. 

Prop.  13.  The  tangents  to  a  circle  at  the  extremities  of 
any  chord,  contain  an  angle  which  is  twice  the  angle  contain- 
ed by  the  same  chord  and  a  diameter  drawn  from  either  of  the 
extremities. 

Prop.  14.  Tf  two  circles  cut  each  other,  and  if  from  any 
point  in  the  straight  line  produced  wrhich  joins  their  intersec- 
tions, two  tangents  be  drawn,  one  to  each  circle,  they  will  be 
equal  to  one  another. 

Prop.  15.  If  from  a  point  without  a  circle,  two  tangents  be 
drawn,  the  straight  line  which  joins  the  points  of  contact  will 
be  bisected  at  right  angles  by  a  line  drawn  from  the  centre  to 
the  point  without  the  circle. 

PROBLEMS. 

Prop.  I .  Inscribe  a  square  in  a  given  right-angled  isosceles 
triangle. 

Prop.  2.  Inscribe  a  circle  in  a  given  rhombus. 

Prop.  3.  Describe  a  circle  whose  circumference  shall 
pass  through  one  angle  and  touch  two  sides  of  a  given 
square. 

Prop.  4.  In  a  given  square,  inscribe  an  equilateral  triangle 
having  its  vertex  in  the  middle  of  a  side  of  the  square. 

Prop.  5.  In  a  given  square,  inscribe  an  equilateral  triangle 
having  its  vertex  in  one  angle  of  the  square , 

Prop.  6.  If  the  sides  of  a  triangle  are  in  the  ratio  of  the 
numbers  2,  4,  and  5,  show  whether  it  will  be  acute-angled  or 
obtuse-angled. 

Prop.  7.  Given  the  area  and  hypothenuse  of  a  right-angled 
triangle,  to  construct  the  triangle. 

Prop.  8.  Bisect  a  triangle  by  a  line  drawn  from  a  given 
point  in  one  of  the  sides. 

Prop.  9.  To  a  circle  of  given  radius,  draw  two  tangents 
which  shall  contain  an  angle  equal  to  a  given  angle. 

Prop  10.  Construct  a  triangle,  having  given  one  side,  the 
angle  opposite  to  it,  and  the  ratio  of  the  other  two'  sides. 

Prop.  11.  Construct  a  triangle,  having  given  the  perimeter 
and  the  angles  of  the  triangle.  * 

Prop.  12.  Upon  a  given  base,  describe  a  right-angled  trian- 
gle, having  given  the  perpendicular  from  the  right  angle  upon 
the  hypothenuse. 

Prop.  13.  Construct  a  triangle,  having  given  one  angle,  a 
side  opposite  to  it,  and  the  sum  of  the  other  two  sides. 

Prop.  14.  Construct  a  triangle,  having  given  one  angle,  an 
adjacent  side,  and  the  sum  of  the  other  two  sides 


GEOMETRICAL    EXERCISES.  233 

Prop.  15.  Trisect  a  given  straight  line,  and  hence  divide 
an  equilateral  triangle  into  nine  equal  parts. 

GEOMETRICAL  EXERCISES  ON  BOOK  VI. 
THEOREMS. 

Prop.  1.  The  square  inscribed  in  a  circle  is  equal  to  half 
the  square  described  about  the  same  circle. 

Prop.  2.  Any  number  of  triangles  having  the  same  base  and 
the  same  vertical  angle,  may  be  circumscribed  by  one  circle. 

Prop.  3.  If  an  equilateral  triangle  be  inscribed  in  a  circle, 
each  of  its  sides  will  cut  off  one  fourth  part  of  the  diameter 
drawn  through  the  opposite  angle. 

Prop.  4.  The  circle  inscribed  in  an  equilateral  triangle  has 
the  same  centre  with  the  circle  described  about  the  same  tri- 
angle, and  the  diameter  of  one  is  double  that  of  the  other. 

Prop.  5.  If  an  equilateral  triangle  be  inscribed  in  a  circle, 
and  the  arcs  cut  off  by  two  of  its  sides  be  bisected,  the  line 
joining  the  points  of  bisection  will  be  trisected  by  the  sides. 

Prop.  6.  The  side  of  an  equilateral  triangle  inscribed  in  a 
circle  is  to  the  radius,  as  the  square  root  of  three  is  to  unity. 

Prop.  7.  The  sum  of  the  perpendiculars  let  fall  from  any 
point  within  an  equilateral  triangle  upon  the  sides,  is  equal  to 
the  perpendicular  let  fall  from  one  of  the  angles  upon  the  op- 
posite side. 

Prop.  8.  If  two  circles  be  described,  one  without  and  the 
other  within  a  right-angled  triangle,  the  sum  of  their  diame- 
ters will  be  equal  to  the  sum  of  the  sides  containing  the  right 
angle. 

Prop.  9.  If  a  circle  be  inscribed  in  a  right-angled  triangle, 
the  sum  of  the  t\vo  sides  containing  the  right  angle  will  ex- 
ceed the  hypothenuse,  by  a  line  equal  to  the  diameter  of  the 
inscribed  circle.. 

Prop.  10.  The  square  inscribed  in  a  semicircle  is  to  the 
square  inscribed  in  the  entire  circle,  as  2  to  5. 

Prop.  11.  The  square  inscribed  in  a  semicircle  is  to  the 
square  inscribed  in  a  quadrant  of  the  same  circle,  as  8  to  5. 

Prop.  12.  The  area  of  an  equilateral  triangle  inscribed  in  a 
circle  is  equal  to  half  that  of  the  regular  hexagon  inscribed  in 
the  same  circle. 

Prop.  13.  The  square  of  the  side  of  an  equilateral  triangle 
inscribed  in  a  circle  is  triple  the  square  of  the  side  of  the  reg- 
ular hexagon  inscribed  in  the  same  circle. 

Prop.  14.  The  area  of  a  regular  hexagon  inscribed  in  a 
circle  is  three  fourths  of  the  regular  hexagon  circumscribed 
about  the  same  circle. 


234  GEOMETRICAL    EXERCISES. 

Prop.  15.  The  triangle,  square,  and  hexagon  are  the  only 
regular  polygons  by  which  the  space  about  a  point  can  be 
completely  filled  up. 

frop.  16.  The  perpendiculars  let  fall  from  the  three  an- 
gles of  any  triangle  upon  the  opposite  sides,  intersect  each 
other  in  the  same  point. 

PROBLEMS. 

Prop.  1.  Trisect  a  given  circle  by  dividing  it  into  three 
equal  sectors. 

Prop.  2.  The  centre  of  a  circle  being  given,  find  two  op- 
posite points  in  the  circumference  by  means  of  a  pair  of  com- 
passes only. 

Prop.  3.  Divide  a  right  angle  into  five  equal  parts. 

Prop.  4.  Inscribe  a  square  in  a  given  segment  of  a  circle. 

Prop.  5.  Having  given  the  difference  between  the  diagonal 
and  side  of  a  square,  describe  the  square. 

Prop.  6.  Inscribe  a  square  in  a  given  quadrant. 

Prop.  7.  Inscribe  a  circle  in  a  given  quadrant. 

Prop.  8.  Describe  a  circle  touching  three  given  straight 
lines. 

Prop.  9.  Within  a  given  circle  describe  six  equal  circles, 
touching  each  other  and  also  the  given  circle,  and  show  that 
the  interior  circle  which  touches  them  all,  is  equal  to  each  of 
them. 

Prop.  10.  Within  a  given  circle  describe  eight  equal  cir- 
cles, touching  each  other  and  the  given  circle. 

Prop.  11.  Inscribe  a  regular  hexagon  in  a  given  equilateral 
fai  angle. 

Prop.  12.  Upon  a  given  straight  line  describe  a  regular 
octagon. 


THE    END. 


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